#### Chapter- 7 : Permutations and Combinations

Exercise- 7.1

Q-1: Form 3- digit numbers by using digits 4, 5, 6, 7 and 8 and assume that

(i) The digits can be repeated.

(ii) Digits can’t be repeated in the 3- digit number.

Solution: Based on the formulae given in Permutations and Combinations

(i) There are multiple numbers of ways to form a 3- digit number by using 5 digits as there are ways of filling 3 vacant bottles with balls in succession by the given digits.

As per the questions demand, the repetition of digits is allowed. So, the unit places should be filled by any of the five digits given in the question.

Similarly,

Tens and hundreds places are filled in by any of the five digits given in the question.

Therefore we get, by the principle of multiplication, by using the given 5 digits i.e., 4, 5, 6, 7, and 8, Total number of 3- digit numbers formed is 5 × 5 × 5 = 125.

(ii) As per the questions demand, the repetition of the digits are not allowed. Here, if the units place if filled before other places such as tens and hundreds, then it can be filled by any of the five digits i.e., 4, 5, 6, 7 and 8.

As the digits can’t be repeated and we have 5 digits, then the total ways of filling the ones place of the 3- digit number is 5.

Now we know that, remaining numbers that we can use at tens place is 4. Now we know that, at tens place, we can use the rest of the 4 digits from the 5 digits given in the question. Similarly, at hundreds place, we can use rest of the 3 digits from the 5- digits given in the question except, that which is already filled at ones place.

By using the principle of multiplication,

The total number of ways to form a three- digits numbers without repetition of the given digits is 5 × 4 × 3 = 60.

Q-2: Form a 3- digit number which should be even by using 1, 2, 3, 6, 8, and 9. Note that, repetition of digits is allowed.

Solution: Based on the formulae given in Permutations and Combinations

There are multiple numbers of ways to form a 3- digit number by using 6 digits, i.e., 1, 2, 4, 6, 8 and 9, as the way we fill 3 vacant bottles with balls in succession by the given balls.

The number should be an even number. So, the unit places should be filled by either 2 or 6 or 8, i.e., we can fill the ones place in 3 ways.

As per the question demand, the repetition of digits is allowed. So, the tens place should be filled by either of the given 6 digits, i.e., 1, 2, 4, 6, 8 and 9 and the hundreds place should also be filled by either of the 6 digits i.e., 1, 2, 4, 6, 8 and 9 in 6 different ways.

Therefore we get, by the principle of multiplication, by using the given 6 digits i.e., i.e., 1, 2, 4, 6, 8 and 9, Total of 3- digit even numbers formed is 3 × 6 × 6 = 108.

Q-3: Find the total number of 4- letter code formed by using the 10 English alphabets, i.e., a, b, c, d, e, f, g, h, i and j. Assume that the repetition of the letters of English alphabet is not allowed.

Solution: Based on the formulae given in Permutations and Combinations

There are multiple numbers of ways to form a 4- letter code ;by using 10 letters from the English alphabet, i.e., a, b, c, d, e, f, g, h, i and j, as the way we fill 4 vacant bottles with balls in succession by the given balls.

Repetition of the letters from English alphabet is not allowed.

So,

If we start filling up the letters from ones place, then we can fill it in 10 different ways as we have 10 letters from the English alphabets.

Now we know that, we can fill tens place in 9 different ways from the remaining 9 letters from the 9 English alphabets, and also on hundreds place we can fill it by 8 different ways from the remaining 8 letters from the English alphabets, and at hundreds place, we can fill it in 7 different ways from the remaining 7 letters from the English alphabets as the repetition is restricted.

Hence, by the principle of multiplication, by using the given 10 letters from the English alphabets, i.e., a, b, c, d, e, f, g, h, i and j, total of 4- letter code formed is 10 × 9 × 8 × 7 = 5040.

Therefore we get, 5040 four- letter code is formed by using the given 10 letters i.e., a, b, c, d, e, f, g, h, i and j from the English alphabets, and there is no repetition in the code hence formed.

Q-4: Find the total number of 5- digit telephone numbers formed by using the numbers from 0 to 9. Note that, each of the telephone number will start with 36 and the repetition of the numbers is not allowed.

Solution:Based on the formulae given in Permutations and Combinations

In the question, it is given that the telephone number is of 5 digits and should start with 36.

So, there will be several telephone numbers as we have to get 5- digit number where 2 digit is fixed that is the first two numbers will be 3 and 6, respectively, i.e., 3, 6, _, _, _ by using the digits from 0 to 9, except the digits 3 and 6.

If we start filling up the numbers to form a telephone number, from ones place then we can fill it in 8 different ways as we have 8 digits from the 0 – 9.

Now we know that, we can fill tens place in 7 different ways from the remaining 7 digits from the 0 to 9, and also on hundreds place we can fill it in 6 different ways from the remaining 6 digits from the 0 to 9, as the repetition is restricted.

Hence, by the principle of multiplication, the total number of ways to form 5- digit telephone number is 8 × 7 × 6 = 336.

Q-5: Consider a scenario when a person tosses a coin. He tosses the coin for two times and every time the outcome of the toss is recorded by one of his friend. Find the total number of possible outcomes of the coin his friend will record.

Solution:Based on the formulae given in Permutations and Combinations

A person tossed a coin for two times and recorded the outcome every time.

Whenever the coin will be tossed, the number of outcome of the coin is 2 which are usually called as a head or a tail or we can say that, in each throw, the different ways to show a different face is 2.

Therefore we get, by the principle of multiplication, the total number of the outcomes possible by the throw can be 2 × 2 = 4.

Q-6: Consider 5 flags which are of different colors. Find the total possibility of getting different signals. Note that, each of the signals requires the use of any two flags at the same time, one after the other.

Solution:Based on the formulae given in Permutations and Combinations

As per the scenario given in the question, each of the signals uses 2 flags at a time.

There are so many possibilities of the combination as the number of ways of filling the 2 vacant boxes in succession of the given 5 flags which are of different color.

Now we know that, let us start with the upper portion of the signal be filled with 5 different ways with any of the 5 flags. After picking up one, the lower portion of the signal will be followed by the upper one, which must be filled up with 4 different signals by any one of the 4 different colored flags left.

Therefore we get, by the principle of multiplication, the total number of the signals which will be generated by 5 different colors of flag is 5 × 4 = 20.

EXERCISE- 7.2

Q-1: Evaluate:

(i). 8! (ii). 4! – 3!

Solution:Based on the formulae given in Permutations and Combinations

(i). 8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320

(ii). 4! = 1 × 2 × 3 × 4 = 24

And, 3! = 1 × 2 × 3 = 6

Therefore we get, 4! – 3! = 24 – 6 = 18

Q-2: Is 3! + 4! = 7! ?

Solution:Based on the formulae given in Permutations and Combinations

Now we know that, 3! = 1 × 2 × 3 = 6

4! = 1 × 2 × 3 × 4 = 24

And, 7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040

Therefore we get, 4! – 3! ≠ 7!

Q-3: Compute:

$$\frac{ 9! }{ 5!\; \times \;3! }$$

Solution:Based on the formulae given in Permutations and Combinations

$$\frac{ 9! }{ 5!\; \times \;3! }$$ = $$\frac{ 1 \;\times \;2 \;\times \;3 \;\times \;4\; \times \;5\; \times \;6\; \times \;7\; \times \;8\; \times \;9 }{ 1\; \times \;2\; \times \;3\; \times \;4\; \times \;5\; \times \;1\; \times \;2\; \times \;3 }\\$$

Therefore we get, $$\frac{ 9! }{ 5! \;\times \;3! }$$ = 392

Q-4. If $$\frac{ 1 }{ 5! } + \frac{ 1 }{ 6! } = \frac{ a }{ 7! }$$, find the value of a.

Solution:Based on the formulae given in Permutations and Combinations

$$\;\frac{ 1 }{ 5! } + \frac{ 1 }{ 6! } = \frac{ a }{ 7! }\\$$

$$\\\Rightarrow$$ $$\frac{ 1 }{ 5! } + \frac{ 1 }{ 6 \;\times \;5! } = \frac{ a }{ 7 \;\times \;6 \times \;5! }\\$$

$$\\\Rightarrow$$ $$\frac{ 1 }{ 5! } + \left [ 1 + \frac{ 1 }{ 6 } \right ] = \frac{ a }{ 7 \;\times \;6 \times \;5! }\\$$

$$\\\Rightarrow 1 + \frac{ 1 }{ 6 } = \frac{ a}{ 7 \;\times \;6 }\\$$

$$\\\Rightarrow$$ $$\frac{ 7 }{ 6 } = \frac{ a }{ 7 \times 6 }\\$$

$$\Rightarrow a = \frac{ 7 \;\times 7 \;\times 6 }{ 6 }\\$$

Therefore we get, a = 49

Q-5: Evaluate: $$\frac{ a! }{ \left ( a – r \right )! }$$, when

(i). a = 5, r = 1 (ii). a = 8, r = 4

Solution:Based on the formulae given in Permutations and Combinations

(i) a = 5, r = 1: [ Given ]

$$\\\frac{ a! }{ \left ( a\; – \;r \right )! }$$ = $$\frac{ 5! }{ \left ( 5 \;-\; 1 \right )! }$$= $$\frac{ 5! }{ 4! }$$= $$\frac{ 5\; \times \;4! }{ 4! }\\$$

Therefore we get, $$\frac{ a! }{ \left ( a \;-\; r \right )! }\;$$ = 5

(ii) $$\;\frac{ a! }{ \left ( a \;-\; r \right )! }$$ = $$\frac{ 8! }{ \left ( 8 \;-\; 4 \right )! }$$= $$\frac{ 8! }{ 4! }\\$$

= $$\\\frac{ 8\; \times \;7 \times 6 \;\times \;5 \times \;4! }{ 4! }\\$$= 8 × 7 × 6 × 5 = 15120

Therefore we get, $$\\\frac{ a! }{ \left ( a \;-\; r \right )! }$$ = 15120

EXERCISE – 7.3

Q-1: If there is no repetition of numbers, then find the total of 3- digit numbers formed by using digits from 2 to 9.

Solution:Based on the formulae given in Permutations and Combinations

Digits in between 2 to 9 are 8.

We have to construct 3- digit numbers by using the digits from 2 to 9.

It is given that; the repetition of digit is restricted.

So, from the permutation point, there are permutations of 8 digits ( different ) among which we need to take 3 at a time.

Hence, the total of 3- digit number formed by using digits from 2 to 9 = $$^{ 8 }P_{ 3 } = \frac{ 8! }{ \left ( 8\; – \;3 \right )!} = \frac{ 8! }{ 5! }$$

= $$\frac{ 8\; \times \;7\; \times \;6\; \times 5! }{ 5! }$$ = 336

Therefore we get, total of 336 3-digit numbers can be formed by using digits from 2 to 9 without any repetition.

Q-2: Find the total number of 4- digits numbers where there is no repetition of digits.

Solution:Based on the formulae given in Permutations and Combinations

We have to construct 4- digit numbers from 0 to 9.

Repetition of digit is restricted.

We know that to have a proper number the thousands place must be filled with 9 digits from 1 to 9, because 0 will be considered as negligible at thousands place.

Therefore we get, the total number of ways through which the thousands place should be filled is 9.

Now we know that, the hundreds place must be filled with remaining 9 digits between 0 to 9, except that number which is inserted at thousands place, as the repetition is restricted. Similarly, tens and ones place can be filled by the remaining 9 digits.

Hence, for hundreds, tens and ones, there will be 3- digit number and therefore, the permutations of 9 different digits will be taken 3 at the same time.

Now we know that,

The total number of 3- digit number:

= $$^{ 9 }P_{ 3 } = \frac{ 9! }{\left ( 9\; – \;3 \right )!} = \frac{ 9! }{ 6! }$$

= $$\frac{ 9\; \times \;8\; \times \;7\; \times \;6! }{ 6! }$$ = 9 × 8 × 7 = 504

Therefore we get, by the principle of the multiplication, the total number of the 4 – digit number is 504 × 9 = 4536

Q-3: Find the total number of 3- digit number which is even and is formed by using the digits 2, 3, 4, 5, 6, 7 and 8. Note that repetition of digit is restricted.

Solution:Based on the formulae given in Permutations and Combinations

We need to form even numbers of 3- digits which can be formed by using the digits such as 2, 3, 4, 5, 6, 7 and 8. Repetition of the digit is restricted.

To make the 3- digit number even, the unit place must be filled with an even number that is 2, 4, 6, and 8 among the given number.

As it is given in the question that the repetition of the numbers is restricted so, the unit place must be already occupied with an even number, and the hundreds and tens place is to be filled with the remaining 6 digits given in the question.

Hence, to fill up the hundred and tens place we need to pick up digits except that which is being filled on ones place. So, we need to pick with permutation of 6 different digits among which 2 digits are taken at a time.

Hence, the number of ways through which the hundreds and tens place is filled:

= $$^{ 6 }P_{ 2 } = \frac{ 6! }{\left ( 6\; -\; 2 \right )!} = \frac{ 9! }{ 4! }$$

= $$\frac{ 6\; \times \;5\; \times 4! }{ 4! }$$ = 30

Therefore we get, by the principle of the multiplication, the number of the 3- digit numbers required is 4 × 30 = 120.

Q-4: How many 4- digit numbers will be formed by using the digits 2, 3, 4, 5 and 6. Note that, repetition of digits is not allowed. Find among those numbers, how many numbers will be even number?

Solution:Based on the formulae given in Permutations and Combinations

We need to construct 4 – digit numbers using the digits 2, 3, 4, 5 and 6.

We have 5 digits among which we need to form 4- digit number as there are permutations of 5 different digits taken 4 at a time.

Hence, the required number of the 4- digit numbers:

=$$^{ 5 }P_{ 4 } = \frac{ 5! }{\left ( 5 \;-\; 4 \right )!} = \frac{ 5! }{ 1! }$$

= $$\frac{ 5\; \times \;4\; \times \;3\; \times \;2\; \times 1 }{ 1 }$$ = 120

Among these 120 numbers of 4 – digit which is formed by using 2, 3, 4, 5 and 6, even numbers must end with an even number which is 2, 4, or 6.

There are chances that there are 3! Number of ways through which the ones place is filled with digits, i.e., it can be filled by 6 number of ways.

As, in the question, it is mentioned that repetition of digits are restricted and the units place is already occupied with an even number, so the remaining place is to be filled by the remaining 4 digits which is not inserted at any place.

Hence, the number of ways by which the thousands, hundreds and tens place will be filled is the permutation of 4 different digits which is taken 3 at a time.

Therefore we get, the number of ways to fill the remaining places:

= $$^{ 4 }P_{ 3 } = \frac{ 4! }{\left ( 4\; -\; 3 \right )!} = \frac{ 4! }{ 1! }$$

= $$\frac{ 4\; \times \;3 \times \;2\; \times \;1 }{ 1 }$$= 24

Hence, by the principle of multiplication, the required number which is an even number is 24 × 3 = 72.

Q-5: Consider a committee with 9 persons, in how many ways can we choose a chairman and a vice chairman? Note that, a person can hold only one position at a time.

Solution:Based on the formulae given in Permutations and Combinations

There is a committee with 9 persons among which, a vice- chairman and a chairman is chosen in such a way that one position can be held by only one person.

So,

To choose a chairman and a vice- chairman, there are number of ways with permutation of 8 different persons taken 2 at a time.

Therefore we get, the number of ways required:

= $$^{ 9 }P_{ 2 } = \frac{ 9! }{\left ( 9\; – \;2 \right )!} = \frac{ 9! }{ 7! }$$

= $$\frac{ 9\; \times \;8 \times \;7! }{ 7! }$$= 72

Q-6: Find n, if $$^{ n \;-\; 1 }P_{ 3 } : \;^{ n }P_{ 4 }$$ = 1 : 9.

Solution:Based on the formulae given in Permutations and Combinations

$$^{ n \;-\; 1 }P_{ 3 } : ^{ n }P_{ 4 } = 1 : 9\\$$

$$\\\Rightarrow$$ $$\frac{^{ n \;- \;1 }P_{ 3 }}{^{ n }P_{ 4 }} = \frac{ 1 }{ 9 }\\$$

$$\\\Rightarrow$$ $$\frac{\left [ \frac{\left ( n\; -\; 1 \right )!}{\left ( n\; – \;1\; – \;3 \right )!} \right ]}{\left [ \frac{ n! }{ \left ( n \;-\; 4 \right )!} \right ]} = \frac{ 1 }{ 9 }$$

$$\\\Rightarrow$$ $$\frac{\left ( n\; – \;1 \right )!}{\left ( n\; -\; 4 \right )!} \times \frac{\left ( n \;-\; 4 \right )!}{ n! } = \frac{ 1 }{ 9 }\\$$

$$\\\Rightarrow$$ $$\frac{\left ( n \;-\; 1 \right )!}{ n \times \left ( n \;- \;1 \right )!} = \frac{ 1 }{ 9 }\\$$

$$\\\Rightarrow \frac{ 1 }{ n } = \frac{ 1 }{ 9 }\\$$

Therefore we get, n = 9

Q-7: Find the value of a, if:

(i). $$^{ 5 }P_{ a } = 2 \;[^{ 6 }P_{ a – 1 }]$$ (ii). $$^{ 5 }P_{ a } = ^{ 6 }P_{ a – 1 }$$

Solution:Based on the formulae given in Permutations and Combinations

(i). $$^{ 5 }P_{ a } = 2 \;[^{ 6 }P_{ a – 1 }]\\$$

$$\\\Rightarrow$$ $$\frac{ 5! }{ \left ( 5 \;-\; a \right )! } = 2 \times \frac{ 6! }{ \left ( 6\; – \;a \;+\; 1 \right )! }\\$$

$$\\\Rightarrow$$ $$\frac{ 5! }{ \left ( 5 – a \right )! } = \frac{ 2 \times 6! }{ \left ( 7 – a \right )! }\\$$

$$\\\Rightarrow$$ $$\frac{ 5! }{ \left ( 5 – a \right )! } = \frac{ 2\; \times \;6\; \times \;5! }{ \left ( 7\; – \;a \right )\;\left ( 6\; -\;a \right )\;\left ( 5\; – \;a \right )! }\\$$

$$\\\Rightarrow$$ $$1 = \frac{ 2\; \times \;6 }{\left ( 7\; – \;a \right )\left ( 6\; – \;a \right )}$$

So, ( 7 – a ) ( 6 – a ) = 12

$$\Rightarrow$$ $$42 – 6a – 7a + a^{ 2 } = 12$$

$$\\\Rightarrow$$ $$a^{ 2 } – 13a + 42 = 12$$

$$\\\Rightarrow$$ $$a^{ 2 } – 13a + 30 = 0$$

$$\\\Rightarrow$$ $$a^{ 2 } –3a – 10a + 30 = 0$$

$$\\\Rightarrow$$ $$a\left ( a – 3 \right ) – 10\left ( a – 3 \right ) = 0$$

$$\\\Rightarrow$$ $$\left ( a – 3 \right ) \left ( a – 10 \right ) = 0\\$$

So, a = 3 or a = 10

It is known that, $$^{ n }P_{ r } = \frac{ n! }{ \left (n\; -\; r \right )! },\; where \; 0 \leq a \leq n$$

$$\Rightarrow 0 \leq a \leq 5\;\\$$

Therefore we get, a ≠ 10

Hence, r = 3

(ii). $$^{ 5 }P_{ a } = \;^{ 6 }P_{ a\; -\; 1 }$$

$$\\\Rightarrow \frac{ 5! }{ \left ( 5\; -\; a \right )! } = \frac{ 6! }{ \left ( 6\; -\; a \;+\; 1 \right )! }\\$$

$$\\\Rightarrow$$ $$\frac{ 5! }{ \left ( 5 \;-\; a \right )! } = \frac{ 6! }{ \left ( 7 \;-\; a \right )! }\\$$

$$\\\Rightarrow$$ $$\frac{ 5! }{ \left ( 5\; -\; a \right )! } = \frac{ 6\; \times \;5! }{ \left ( 7 \;- \;a \right )\;\left ( 6 \;-\; a \right )\;\left ( 5\; -\; a \right )! }\\$$

$$\\\Rightarrow$$ $$1 = \frac{ 6 }{\left ( 7 \;- \;a \right )\;\left ( 6\; -\; a \right )}$$

So, ( 7 – a ) ( 6 – a ) = 6

$$\\\Rightarrow$$ $$42 – 6a – 7a + a^{ 2 } = 6$$

$$\\\Rightarrow$$ $$a^{ 2 } – 13a + 42 = 6$$

$$\\\Rightarrow$$ $$a^{ 2 } – 13a + 36 = 0$$

$$\\\Rightarrow$$ $$a^{ 2 } –4a – 9a + 36 = 0$$

$$\\\Rightarrow$$ $$a\left ( a\; -\; 4 \right ) – 9\left ( a\; -\; 4 \right ) = 0$$

$$\\\Rightarrow$$ $$\left ( a\; -\; 4 \right ) \left ( a\; –\; 9 \right ) = 0$$

So, a = 4 or a = 9

It is known that, $$^{ n }P_{ r } = \frac{ n! }{ \left (n\; – \;r \right )! },\; where \; 0 \leq a \leq n$$

$$\Rightarrow 0 \leq a \leq 5\\$$

Therefore we get, a ≠ 9

Hence, a = 4

Q-8: Get all the number of ways, with or without the meaning, which might be formed by using the letters from the word EQUATION. Note that, repetition of the word is restricted.

Solution:Based on the formulae given in Permutations and Combinations

In the word EQUATION, we have 8 letters from the English alphabets.

Hence, to find the number of words formed by using all those 8 letters of the word EQUATION, without repetition of the alphabets, we have:

Permutation of 8 different letters taken 8 at a time, which is $$^{ 8 }P_{ 8 }$$ = 8!

Hence, the number of words required which can be formed = 8! = 40320.

Q-9: Find the number of words, with or without the meaning, which might be formed by using the letters from the word MONDAY. Note that, repetition of the word is restricted, if

(i) At a time, 3 letters can be used.

(ii) All the letters can be used at a time.

(iii) First letter is a vowel and all the letters are being used.

Solution:Based on the formulae given in Permutations and Combinations

In the word MONDAY, we have 6 letters from the English alphabets.

(i) As per the question demand, we need to use 3 letters at a time.

So, the number of words formed by the letters of the word MONDAY, without the repetition of the letters, is

The permutation of 6 different objects which are taken 3 at a time is given by, $$^{ 6 }P_{ 3 }$$

Hence, the number of words required which can be formed by using the word MONDAY using only 3 letters at a time is given by,

$$^{ 6 }P_{ 3 }$$ = $$\frac{ 6! }{ \left ( 6\; -\; 3 \right )! }$$=$$\frac{ 6! }{ 3! }$$=$$\Rightarrow \frac{ 6\; \times \;5\; \times \;4\; \times \;3! }{ 3! }$$= 6 × 5 × 4 = 120 words

(ii) In this case, at a time, we can use all the letters of the given word MONDAY.

So, to find the number of words formed, we have:

Permutation of 6 different letters taken any of the 6 at a time, that is: $$^{ 6 }P_{ 6 }$$.

Hence, the number of words required which can be formed by using all the letters of the word MONDAY at a time = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.

(iii) In the given words MONDAY, we have two vowels which is O and A.

As per the given condition, vowels must be filled at the start of the word. Also, the repetition of letter is restricted.

So, to this can be done only in 2! Number of ways.

We have 6 letters and the repetition is restricted, also the rightmost place will be filled by a vowel from the word MONDAY, so the remaining 5 places will be filled by the remaining 5 letters. And, this can be done in 5! Number of ways.

Hence, in this case, the number of words required and being formed is 5! × 2! = 5 × 4 × 3 × 2 × 1 × 2 × 1 = 240.

Q-10: Find the number of distinct permutations such that 4 I’s does not come together in the word MISSISSIPPI.

Solution:Based on the formulae given in Permutations and Combinations

In the word MISSISSIPPI, M appears for 1 time, I appears for 4 times, P appears for 4 times and S appears for 4 times.

So, the required number of the distinct permutations of the letters in the word MISSISSIPPI is given by:

$$\frac{ 11! }{ 4! \times 4! \times 2! }$$ = $$\frac{ 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4! }{ 4! \times 4 \times 3 \times 2 \times 1 \times 2 \times 1 }$$= $$\frac{ 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 }{ 4 \times 3 \times 2 \times 1 \times 2 \times 1 }$$= 34650.

As there are 4 I’s in this given number which is MISSISSIPPI, whenever they will occur together, they will be treated as a single object which is IIII. Since, all the 4 I’s are considered as 1 and there are 7 letters left in the given word MISSISSIPPI, so we have total of 8 words in account.

Among these 8 words S appears for 4 times and P appears for 4 times, and M appears just once.

So, this can be arranged in $$\frac{ 8! }{ 4! \;\times \;2! }$$ ways, i.ie., 8 × 7 × 3 × 5 = 840 ways

So, the number of arrangements where all I’s can occur together = 840

Hence, the total number of the distinct permutations of the letters in the given word MISSISSIPPI, where all the 4 I’s won’t come together = 34650 – 840 = 33810.

Q-11: Arrange the letters of the word PERMUTATIONS and observe the number of ways of the permutation if,

(i). the new word starts with P and ends with S.

(ii). when the vowels are taken all together.

(iii). there will always be 4 letters between P and S.

Solution:Based on the formulae given in Permutations and Combinations

In the given word PERMUTATIONS, all the other letters occurred for once except T which occurred for twice.

(i) As we need to fix P and S at the extreme ends i.e., P at the left side of the word and S on the right side of the word, then we have 10 more letters to fix.

Therefore we get, number of arrangements required in this case = $$\frac{ 10! }{ 2! }$$ = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 = 1814400

(ii) In the given word PERMUTATIONS, we have all the 5 vowels, i.e., a, e, i, o, u and each of the vowel is appearing only for once in the given word.

As in this case, they have to occur together, so these vowels are considered as a single object. Since, we have 12 letters in the given word and as all the vowels are counted as one, so we have more 7 letters. So, 8 objects are there in the account.

Now we know that,

All these letters can be arranged in $$\frac{ 8! }{ 2! }$$ ways where there is 2 T’s.

As there are 5 different vowels, then corresponding to these arrangements, vowels will be arranged in 5! Ways.

Hence, by the principle of multiplication, number of arrangements required for this case = $$\frac{ 8! }{ 2! } \times 5!$$ = 2419200.

(iii) We have to arrange the letters of the word PERMUTATIONS in such a way that always there will be 4 letters in between P and S.

So, it is clear that the position of P and S is fixed and 4 others letters can be inserted between P and S. There are 2 T’s in the remaining 10 letters which can be arranged in $$\frac{ 10! }{ 2! }$$ ways.

Also, 4 letters can be placed in between P and S in 2 × 7 = 14 ways.

Hence, by the principle of multiplication, number of arrangements required in this case = $$\frac{ 10! }{ 2! } \times 14$$ = 25401600.

EXERCISE – 7.4

Q-1: If $$^{ n }C_{ 8 } = ^{ n }C_{ 3 }$$, find the value of $$^{ n }C_{ 3 }$$

Solution:Based on the formulae given in Permutations and Combinations

We know that, if $$^{ n }C_{ x } = \;^{ n }C_{ y }$$

So, x = y or n = x + y

Hence, $$^{ n }C_{ 8 } = \;^{ n }C_{ 3 }$$

n = 8 + 3 = 11

Therefore we get, $$^{ n }C_{ 2 } = \;^{ 11 }C_{ 2 }$$=$$\frac{ 11! }{ 2!\left ( 11 \;-\; 2 \right )!}$$ = $$\frac{ 11\; \times \;10\; \times \;9! }{ 2\; \times \;1 \;\times \;9! }$$ = $$\frac{ 11\; \times \;10 }{ 2 }$$

Therefore we get, $$^{ n }C_{ 8 } = \;^{ n }C_{ 3 }$$ = 55

Q-2: Find the value of ‘a’ if:

(i) $$^{ 2a }C_{ 2 } : \;^{ a }C_{ 2 }$$ = 12 : 1

(ii) $$^{ 2a }C_{ 2 } : \;^{ a }C_{ 2 }$$ = 11 : 1

Solution:Based on the formulae given in Permutations and Combinations

(i). $$\frac{ ^{2a }C_{ 2 } }{ ^{ a }C_{ 2 } } = \frac{ 12 }{ 1 }\\$$

$$\\\Rightarrow$$ $$\frac{\left ( 2a \right )! }{ 2!\left ( 2a\; -\; 2 \right )! } \times \frac{ 2!\left ( 2a\; -\; 2 \right )! }{ a! } = \frac{ 12 }{ 1 }\\$$

$$\\\Rightarrow$$ $$\frac{ \left ( 2a \right )\left ( 2a \;- \;1 \right )\left ( 2a \;- \;2 \right )! }{ \left ( 2a\; -\; 2 \right )!} \times \frac{ \left ( a \;-\; 2 \right )! }{ a\left ( a\; – \;1 \right )\left ( a\; – \;2 \right )! } = 12\\$$

$$\\\Rightarrow$$ $$\frac{ \left ( 2 \right )\left ( 2a \;-\; 1 \right ) }{ \left ( a \;-\; 1 \right )\left ( a\; – \;2 \right ) } = 12\\$$

$$\\\Rightarrow$$ $$\frac{ 4 \left ( a \;-\; 1 \right )\left ( 2a \;-\; 1 \right ) }{ \left ( a \;-\; 1 \right )\left ( a \;-\; 2 \right ) } = 12\\$$

$$\\\Rightarrow$$ $$\frac{ \left ( 2a \;- \;1 \right ) }{ \left ( a \;-\; 2 \right ) } = 3$$

So, 2a – 1 = 3( a – 2 )

2a – 1 = 3a – 6

3a – 2a = -1 + 6

Therefore we get, the value of a = 5

(ii). $$\frac{ ^{2a }C_{ 2 } }{ ^{ a }C_{ 2 } } = \frac{ 11 }{ 1 }\\$$

$$\\\Rightarrow$$ $$\frac{\left ( 2a \right )! }{ 2!\left ( 2a\; -\; 2 \right )! } \times \frac{ 2!\left ( 2a \;-\; 2 \right )! }{ a! } = \frac{ 11 }{ 1 }\\$$

$$\\\Rightarrow$$ $$\frac{ \left ( 2a \right )\left ( 2a \;-\; 1 \right )\left ( 2a \;-\; 2 \right )! }{ \left ( 2a\; -\; 2 \right )!} \times \frac{ \left ( a\; -\; 2 \right )! }{ a\left ( a\; -\; 1 \right )\left ( a \;-\; 2 \right )! } = 11\\$$

$$\\\Rightarrow$$ $$\frac{ \left ( 2 \right )\left ( 2a \;- \;1 \right ) }{ \left ( a\; -\; 1 \right )\left ( a \;-\; 2 \right ) } = 11\\$$

$$\\\Rightarrow$$ $$\frac{ 4 \left ( a\; -\; 1 \right )\left ( 2a\; – \;1 \right ) }{ \left ( a\; – \;1 \right )\left ( a \;-\; 2 \right ) } = 11\\$$

$$\\\Rightarrow$$ $$4\left ( 2a \;- \;1 \right ) = 11\left ( a \;-\; 2 \right )$$

So, 8a – 4 = 11( a – 2 )

8a – 4 = 11a – 22

11a – 8a = -4 + 22

3a = 18

Therefore we get, the value of a = 6

Q-3: Find total number of chords which can be drawn through 21 points on the circle.

Solution:Based on the formulae given in Permutations and Combinations

2 points are required to draw one chord on a circle.

Now we know that, in order to obtain the number of chords which can be drawn through the given 21 points on the circle, we need to count total number of the combinations.

Hence, there can be several numbers of chords as there are combinations of 21 points which can be taken 2 at a time.

Therefore we get, number of chords required = $$^{ 21 }C_{ 2 } = \frac{ 21! }{ 2!\left ( 21\; -\; 2 \right )! } = \frac{ 21! }{ 2! \;19! } = \frac{ 21 \;\times \;20 }{ 2 }$$ = 210

Q-4: Find the number of ways in which a team of 2 boys and 2 girls can be selected from the group of 5 boys and 4 girls.

Solution:Based on the formulae given in Permutations and Combinations

We have to select 2 boys and 2 girls from a group of 5 boys and 4 girls to form a team.

Now we know that,

2 boys should be selected among 5 boys group in $$^{ 5 }C_{ 2 }$$ ways.

Also, 2 girls should be selected among a group of 4 girls in $$^{ 4 }C_{ 2 }$$ ways.

Hence, by the principle of multiplication, the number of ways by which a team of 2 boys and 2 girls can be formed:

= $$^{ 5 }C_{ 2 } \times ^{ 4 }C_{ 2 }$$ = $$\frac{ 5! }{ 2! \;3! } \times \frac{ 4! }{ 2! \;2! }$$= $$\frac{ 5\; \times 4\; \times \;3! }{ 2\; \times \;1\; \times \;3! } \times \frac{ 4\; \times \;3\; \times \;2! }{ 2\; \times \;1\; \times \;2! }$$= 5 × 2 × 2 × 3 = 60 ways

Q-5: In how many ways will 9 balls be selected from 7 red balls, 6 blue balls and 5 white balls if every time 3 balls of different colors be selected?

Solution:Based on the formulae given in Permutations and Combinations

We have total of 6 blue balls, 5 white balls and 7 red balls.

We have to select 9 balls from these 18 balls in such a ways that every time 3 different colored balls can be selected.

Here,

We can select 3 balls from 7 red balls in $$^{ 7 }C_{ 3 }$$ number of ways.

We can select 3 balls from 6 blue balls in $$^{ 6 }C_{ 3 }$$ number of ways.

We can select 3 balls from 5 white balls in $$^{ 5 }C_{ 3 }$$ number of ways.

Therefore we get, by the principle of multiplication, number of ways required to select 9 balls:

= $$^{ 7 }C_{ 3 } \times ^{ 6 }C_{ 3 } \times ^{ 5 }C_{ 3 }$$ = $$\frac{ 7! }{ 3!\; 4! } \times \frac{ 6! }{ 3! \;3! } \times \frac{ 5! }{ 3!\; 2! }$$ = $$\frac{ 7\; \times \;6\; \times \;5\; \times \;4! }{ 4!\; \times \;3\; \times \;2\; \times \;1 } \times \frac{ 6\; \times \;5\; \times \;4\; \times \;3! }{ 3!\; \times 3 \;\times 2\; \times \;1 } \times \frac{ 5\; \times \;4\; \times \;3! }{ \;3!\; \times \;2 \;\times \;1 }$$

= 7000 ways

Q-6: Find out the number of combinations of 5 cards from a deck of 52 cards. Assume that there is at least one ace in each of the combinations.

Solution:Based on the formulae given in Permutations and Combinations

We know that, we have 52 cards in a deck where there are 4 aces.

We have to make a combination of 5 cards in which there will be one ace for sure.

As we have 4 aces in the deck so, aces can be selected in $$^{ 4 }C_{ 1 }$$ number of ways and out of 48 cards in the deck, 4 cards can be selected in $$^{ 48 }C_{ 4 }$$ number of ways.

Therefore we get, by the principle of the multiplication, number of the combination of 5 cards required:

= $$^{ 48 }C_{ 4 } \times ^{ 4 }C_{ 1 }$$= $$\frac{ 48! }{ 4!\; 44! } \times \frac{ 4! }{ 1!\; 3! }$$= $$\frac{ 48\; \times 47\; \times \;46\; \times \;45\; \times \;44! }{ 4\; \times 3\; \times 2\; \times 1\; \times \;44! } \times \frac{ 4\; \times \;3! }{ 3! }$$

= $$\frac{ 48\; \times \;47\; \times \;46\; \times \;45 }{ 4\; \times \;3\; \times \;2\; \times \;1 } \times 4$$

= 778320 number of combinations.

Q-7: Find the number of ways by which one can select a cricket team of 11 players from a bunch of 17 players among which only 5 players can bowl if every cricket team of 11 players must have exactly 4 bowlers.

Solution:Based on the formulae given in Permutations and Combinations

Among 17 players, 5 players can bowl.

Now we know that, we need to select a cricket team of 11 players where there are exactly 4 players who can bowl.

So, Among 5, 4 bowlers can be selected in $$^{ 5 }C_{ 4 }$$ number of ways and the remaining 7 players for the team of 11 players will be selected from the remaining 12 players in $$^{ 12 }C_{ 7 }$$ number of ways.

Therefore we get, by the principle of multiplication, number of ways required to select a cricket team:

= $$^{ 5 }C_{ 4 } \times ^{ 12 }C_{ 7 }$$ = $$\frac{ 5! }{ 4!\; 1! } \times \frac{ 12! }{ 7!\; 5! }\\$$

= $$\\\frac{ 5\; \times \;4! }{ 4! } \times \frac{ 12\; \times \;11\; \times \;10\; \times \;9\; \times \;8\; \times \;7! }{ 7!\; \times \;5\; \times \;4\; \times \;3\; \times \;2\; \times \;1 }\\$$

= $$\\\frac{ 5\; \times \;12\; \times \;11\; \times \;10\; \times \;9\; \times \;8 }{ 5\; \times \;4\; \times \;3\; \times \;2\; \times \;1 }\\$$

= 3960 number of ways.

Q-8: There are 6 black and 5 red balls. Find the number of ways through which 3 black and 3 red balls will be selected.

Solution:Based on the formulae given in Permutations and Combinations

We have a bag which contains 6 black and 5 red balls.

3 black balls will be selected from 6 black balls in 6C3 number of ways and 3 red balls will be selected out of 5 red balls in 5C3 number of ways.

Therefore we get, by the principle of multiplication, required number of ways to select 3 black and 3 red balls:

= $$^{ 6 }C_{ 3 } \times ^{ 5 }C_{ 3 }$$= $$\frac{ 6! }{ 3!\; 3! } \times \frac{ 5! }{ 3!\; 2! }$$= $$\frac{ 6\; \times \;5\; \times \;4\; \times \;3! }{ 3!\; \times \;3\; \times \;2\; \times \;1 } \times \frac{ 5\; \times \;4\; \times \;3! }{ 3!\; \times \;2\; \times \;1 }$$= $$5 \times 4 \times 5 \times 2$$

= 200 number of ways.

Q-9: Find the number of ways by which a student can choose a program of 5 courses, if there are options of 10 courses and for every student 2 courses are made compulsory.

Solution:Based on the formulae given in Permutations and Combinations

10 courses are available out of which 2 courses are made compulsory for each student.

Hence, each student have to choose 3 courses from the remaining courses, i.e., 8 other remaining courses, which can be selected in $$^{ 8 }C_{ 3 }$$ number of ways.

Therefore we get, the total number of ways in which program can be chosen:

= $$^{ 8 }C_{ 3 }$$= $$\frac{ 8! }{ 3!\; 5! }$$= $$\frac{ 8\; \times \;7\; \times \;6\; \times \;5! }{ 5!\; \times \;3\; \times \;2\; \times \;1 }$$

= 56 number of ways.

MISCELLANEOUS EXERCISE

Q-1: Find the number of words, having meaning or meaningless, each of 2 vowels and 4 consonants will be formed by the letters of the word DAUGHTER?

Solution:Based on the formulae given in Permutations and Combinations

In DAUGHTER word, we have 3 vowels which is A, U and E, rest 5 letters are consonants which are D, G, H, T and R.

Now we know that,

The number of way to select 2 vowels at a time from 3 vowels will have permutation say, 3C2 = 3

Also, the number of way to select 4 consonants at a time from 5 consonants will have permutation say, 5C4= 5

Hence, the total number of combinations of 2 vowels and 4 consonants = 3 × 5 = 15

These 15 combinations of 2 vowels and 4 consonants will be arranged in 6! number of ways.

Therefore we get, the number of different words required = 15 × 5! = 1800

Q-2: Find the number of words, which can be either meaningful or meaningless, which will be formed by using all of the letters of the word EQUATION at a time such that both the vowel and the consonant will occur together.

Solution:Based on the formulae given in Permutations and Combinations

In the given word EQUATION, we have 5 vowels, that are E, U, A, I and O, and there are 3 consonants, that are Q, T and N.

As all the vowels and consonants will occur together, means both (A, E, I, O, U) and (Q, T, N) will be considered as a single letter.

Thus, the permutation of vowels and consonants taken all at a single time is counted which is: 2P2 = 2!

Now we know that, at each of the permutation for vowels, there is a permutation of 5! Taken all at a time and for consonants, the permutation is 3! taken all at a time.

Therefore we get, by the principle of multiplication, number of the words required = 2! × 5! × 3! = 1440

Q-3: There are 10 boys and 5 girls through which a committee with 7 members has to be formed. Now we know that, find the number of ways in which this might be created when the committee formed consists of:

(i) Exactly 4 girls (ii) at least 4 girls (iii) almost 4 girls?

Solution:Based on the formulae given in Permutations and Combinations

(i) From 10 boys and 5 girls, a committee of 7 members is being formed.

As in this case, exactly 4 girls is needed to be there in the committee, so each committee should contain ( 7 – 4 ) = 3 boys in the committee.

Therefore we get, number of ways required in this case:

= 5C4 × 10C3

= $$\frac{ 5! }{ 4!\; 1! } \times \frac{ 10! }{ 3!\; 7! }$$ = $$5 \times \frac{ 10\; \times \;9\; \times \;8\; \times \;7! }{ 7!\; \times \;3\; \times \;2\; \times \;1 }$$

= 600 number of ways

(ii) Here, in this case, at least 4 girls can be there in each committee.

So, in such cases, a committee can consist of

(a) Either 3 boys or 4 girls (b) either 2 boys or 5 girls.

(a) 4 boys and 3 girls will be selected in 10C3 × 5C4 number of ways.

(b) 2 boys and 5 girls will be selected in 10C2 × 5C5 number of ways.

Hence, the number of ways required in this case:

= 10C3 × 5C4 + 10C2 × 5C5

=$$\frac{ 10! }{ 3!\; 7!} \times \frac{ 5! }{ 4! \;1! } + \frac{ 10! }{ 2! \;8! } \times \frac{ 5! }{ 5! \;0!}$$

= $$\left [\frac{ 10\; \times \;9\; \times \;8\; \times\; 7! }{ 3\; \times \;2\; \times \;1\; \times \;7! } \times \;5\; \right ] + \left [ \frac{ 10\; \times \;9\; \times \;8! }{ 2\; \times \;1\; \times \;8! } \times 1 \right ]$$= ( 10 × 3 × 4 × 5 ) + ( 5 × 9 )

= 645 number of ways.

(iii) Here, in this case, almost 4 girls can be there in each of the committee.

So, in this case, a committee can consist of:

(a) 4 girls and 3 boys (b) 3 girls and 4 boys

(c) 2 girls and 5 boys (d) 1 girl and 6 boys

(e) No girl and 7 boys

(a) 4 girls and 3 boys will be selected in 5C4 × 10C3 number of ways.

(b) 3 girls and 4 boys will be selected in 5C3 × 10C4 number of ways.

(c) 2 girls and 5 boys will be selected in 5C2 × 10C5 number of ways.

(d) 1 girl and 6 boys will be selected in 5C1 × 10C6 number of ways.

(e) No girl and 7 boys will be selected in 5C0 × 10C7 number of ways

Hence, the number of ways required in this case:

= 5C4 × 10C3 + 5C3 × 10C4 + 5C2 × 10C5 + 5C1 × 10C6 + 5C0 × 10C7

= $$\frac{ 5! }{ 4! \;1!} \times \frac{ 10! }{ 3! \;7! } + \frac{ 5! }{ 3! \;2! } \times \frac{ 10! }{ 4!\; 6!} + \frac{ 5! }{ 2!\; 3! } \times \frac{ 10! }{ 5! \;5!} + \frac{ 5! }{ 1!\; 4! } \times \frac{ 10! }{ 6!\; 4!} + \frac{ 5! }{ 0! \;5! } \times \frac{ 10! }{ 7! \;3!}$$

= 300 + 2100 + 2520 + 1050 + 120

= 6090 number of ways.

Q-4: In a dictionary, the permutation of the letters of the word EXAMINATION is listed, then find the total number of words in the list of dictionary before the first letter of the word start with E.

Solution:Based on the formulae given in Permutations and Combinations

There are 11 letters in the given word EXAMINATION where A, I and N appeared twice and all other letter appeared only for once.

Words which are listed before the word which is starting with E in a dictionary can be the word which starts with only A.

Hence, in order to get the numbers of words starting from A, means the letter A will be fixed at the extreme left position of the word, and there are 10 remaining letters taken at a time is rearranged.

In the given word which is EXAMINATION, there are 2 N’s and 2 I’s in the remaining letters and other letters occurred only for once.

Thus, the number of words starting with the letter A:

= $$\frac{ 10! }{ 2!\; 2! }$$ = 907200

Hence, the number of words required is 907200.

Q-5: Find the total number of 6- digit numbers formed by using the digits 0, 2, 3, 4, 5 and 6 which should be divisible by 10. Note that, the repetition of digit is not allowed.

Solution:Based on the formulae given in Permutations and Combinations

For a number to be divisible by 10, then its unit place must be filled with 0 only.

Thus, units place is fixed which is to be filled only by 0.

Hence, there are multiple numbers of ways to fill up the other 5 vacant places in succession with the other digits that is, 2, 3, 4, 5 and 6.

Now we know that,

Other 5 vacant places will be filled in by 5! number of ways.

Therefore we get, the number of 6-digit numbers required = 5! = 120

Q-6: It is known that the English alphabet has 21 consonants and 5 vowels. Find the number of words formed with 4 different consonants and 3 different vowels from the English alphabet.

Solution:Based on the formulae given in Permutations and Combinations

We need to select 4 different consonants from 21 consonants and 3 different vowels from 5 vowels of the English alphabet.

So, the permutation of the selection of vowels is given by:

5C3 = $$\frac{ 5! }{ 3!\; 2! }$$ = 10

Also, the permutation of the selection of consonants is given by:

21C2= $$\frac{ 21! }{ 4! \;17! }$$ = 3990

Hence, the total number of the combinations of 4 different consonants and 3 different consonants = 10 × 3990 = 39900.

Q-7: There are 12 questions in the question paper of an examination which is mainly divided into two parts, say, part-I and part-II, each having 4 and 8 questions, respectively. There is a condition given in the question paper that the student has to attempt at least 8 questions for sure, selection 3 from each section. Now we know that, find the number of ways in which a student can select the questions in the question paper.

Solution:Based on the formulae given in Permutations and Combinations

From the data given in the question, know that the examination question paper is divided into two parts, namely; part-I and part-II, containing 4 and 8 questions each.

A student need to attempt at least 8 questions, selecting 3 questions at least from each sections, which can be done in following ways:

(a) 4 questions from part I and 4 questions from part II

(b) 3 questions from part I and 5 questions from part II

The selection of 4 questions from part I and 4 questions from part II have permutation as: 4C4 × 8C4 number of ways.

The selection of 3 questions from part I and 5 questions from part II have permutation as: 4C3 × 8C5 number of ways.

Now we know that,

The total number of ways in which selection of questions are required:

= 4C4 × 8C4 + 4C3 × 8C5

=$$\frac{ 4! }{ 4!\; 0! } \times \frac{ 8! }{ 4!\; 4! }$$ + $$\frac{ 4! }{ 3!\; 1! } \times \frac{ 8! }{ 5! \;3! }$$ = $$1 + \frac{ 8\; \times \;7\; \times \;6\; \times \;5! }{ 5!\; \times \;3\; \times \;2\; \times \;1 }$$

= 57 number of ways

Q-8: Find the number of ways of 5- card combinations from a deck of 52 cards. Note that, among those 5 cards, there must be exactly one king in the combination.

Solution:Based on the formulae given in Permutations and Combinations

We need to make 5 – card combinations from a deck of 52 cards, in such a way that there must be a king in the combination.

We know that, there are 4 kings in a deck of 52 cards.

Then,

The permutation of the selection of a king from the deck is: 4C1 number of ways.

Now we know that after selection of a king for the combination of 5 cards then, there are 4 cards left to be selected.

Thus, 4 cards can be selected from the deck with 48 cards left in 48C4 number of ways.

Therefore we get, the number of 5- card combination required is:

=4C1 × 48C4 = $$\frac{ 4! }{ 1!\; 3! } \times \frac{ 48! }{ 4! \;44! }$$ = $$\frac{ 48\; \times \;47\; \times \;46\; \times\;45\; \times \;44! }{ 3\; \times \;2\; \times \;1\; \times \;44! }$$

= 7,78,320 number of combinations.

Q-9: Consider a situation in which 7 men and 6 women needs to be seated in a row in such a way that the women will occupy the even places.

Solution:Based on the formulae given in Permutations and Combinations

It is required to arrange the seating of 7 men and 6 women in a row in such a way that the women will occupy the even places.

7 men can be seated in 7! Ways. For every arrangement, 6 women will be seated only at the blank ( _ ) places so that the women can occupy the even places.

A _ A _ A _ A _ A _ A_ A

Thus, the women will be seated in 6! number of ways.

Therefore we get, Number of arrangements possible as per the given condition = 6! × 7! = 3628800 number of arrangements.

Q-10: There are 20 students in a class from which 8 students will be chosen for an exclusion party. 3 students from the class decided that either they will join combinely or none of them will go for the exclusion party. Find the number of ways by which the exclusion party will be chosen.

Solution:Based on the formulae given in Permutations and Combinations

8 students will be selected for the exclusion party from a class of 20 students.

Also, it is given in the question that, 3 students from that class decided that they will either go together or none of them will go for the exclusion party. Thus, there are mainly 2 cases:

(a) All 3 of them are going for the exclusion party.

Thus, the remaining 5 students will be selected from the remaining 17 students in the class in $$^{ 17 }C_{ 5 }$$ number of ways.

(b) None of them are going for the exclusion party.

Thus, 8 students will be selected from the class of 20 students in $$^{ 20 }C_{ 8 }$$ number of ways.

Therefore we get, the number of ways required for choosing students for the exclusion party is:

=$$^{ 17 }C_{ 5 } + \;^{ 20 }C_{ 8 }$$ = $$\frac{ 17! }{ 5! \;12! } \times \frac{ 20! }{ 8! \;12! }\\$$

= $$\frac{ 17\; \times \;16\; \times \;15\; \times \;14\; \times \;13\; \times \;12! }{ 5\; \times \;4\; \times \;3\; \times \;2\; \times \;1\; \times \;12! }$$ + $$\frac{ 20\; \times \;19\; \times \;18\; \times \;17\; \times \;16\; \times \;15\; \times \;14\; \times \;13\; \times \;12! }{ 8\; \times \;7\; \times \;6\; \times \;5\; \times \;4\; \times \;3\; \times \;2\; \times \;1\; \times 12! }$$ = 6188 + 125970

= 132158 number of ways to choose the students for the exclusion party.

Q-11: Arrange ASSASSINATION in such a way that all the S’s are filled together. Find the number ways for the arrangement of S’s together.

Solution:Based on the formulae given in Permutations and Combinations

ASSASSINATION is the word given in the question where A appeared for 3 times, S appeared for 4 times, I appeared for 2 times, N appeared for 2 times and, T and O appeared just for once.

We need to arrange all the words in such a way that the all S’s are placed together.

Now we know that,

SSSS is considered as a single object.

Since, including the single object we have remaining 9 letters, so we have the count of 10 letters more.

Thus, the 10 letters where they have 3 A’s, 2 I’s and 2 N’s will be arranged in $$\frac{ 10! }{ 3! \;2! \;2! }$$ number of ways.

Therefore we get, the total number of ways required to arrange the letters of ASSASSINATION word = $$\frac{ 10! }{ 3! \;2! \;2! }$$ = 151200