Factorial Notation -> ! ; n! = n(n – 1) (n – 2) ... 3.2.1 = Product of n consecutive integers starting from 1.
0! = 1
Factorials of only Natural numbers are defined. n! is defined only for n ≥ 0 n! is not defined for n < 0
^{n}C_{r} = 1 when n = r.
Combinations (represented by ^{n}C_{r}) can be defined as the number of ways in which r things at a time can be SELECTED from amongst n things available for selection. The key word here is SELECTION.
Understand here that the order in which the r things are selected has no importance in the counting of combinations.
^{n}C_{r} = Number of combinations (selections) of n things taken r at a time. ^{n}C_{r} = n! /[r! (n–r)!] ; where n ≥ r (n is greater than or equal to r).
Some typical situations where selection/combination is used:
Selection of people for a team, a party, a job, an office etc. (e.g. Selection of a cricket team of 11 from 16 members)
Selection of a set of objects (like letters, hats, points pants, shirts, etc) from amongst another set available for selection.
In other words any selection in which the order of selection holds no importance is counted by using combinations.
Permutations (represented by ^{n}P_{r}) can be defined as the number of ways in which r things at a time can be SELECTED & ARRANGED at a time from amongst n things.
The key word here is ARRANGEMENT. Hence please understand here that the order in which the r things are arranged has critical importance in the counting of permutations.
In other words permutations can also be referred to as an ORDERED SELECTION. ^{n}P_{r} = number of permutations (arrangements) of n things taken r at a time.
^{n}P_{r} = n!/ (n – r)!; n ≥ r
Some typical situations where ordered selection/ permutations are used:
Making words and numbers from a set of available letters and digits respectively
Filling posts with people
Selection of batting order of a cricket team of 11 from 16 members
Putting distinct objects/people in distinct places, e.g. making people sit, putting letters in envelopes, finishing order in horse race, etc.)
Selection : Suppose we have three men A, B and C out of which 2 men have to be selected to two posts. This can be done in the following ways: AB, AC or BC (These three represent the basic selections of 2 people out of three which are possible. Physically they can be counted as 3 distinct selections. This value can also be got by using ^{3}C_{2}. Note here that we are counting AB and BA as one single selection. So also AC and CA and BC and CB are considered to be the same instances of selection since the order of selection is not important.
Arrangement : Suppose we have three men A, B and C out of which 2 men have to be selected to the post of captain and vice captain of a team. In this case we have to take AB and BA as two different instances since the order of the arrangement makes a difference in who is the captain and who is the vice captain. Similarly, we have BC and CB and AC and CA as 4 more instances. Thus in all there could be 6 arrangements of 2 things out of three. This is given by 3P2 = 6.
When we look at the formulae for Permutations and Combinations and compare the two we see that, nPr = r! × nCr = nCr × rPr
This in words can be said as: The permutation or arrangement of r things out of n is nothing but the selection of r things out of n followed by the arrangement of the r selected things amongst themselves
MNP Rule : If there are three things to do and there are M ways of doing the first thing, N ways of doing the second thing and P ways of doing the third thing then there will be M × N × P ways of doing all the three things together.The works are mutually inclusive.
This is used to for situations like: The numbers 1, 2, 3, 4 and 5 are to be used for forming 3 digit numbers without repetition. In how many ways can this be done?
Using the MNP rule you can visualise this as: There are three things to doÆ The first digit can be selected in 5 distinct ways, the second can be selected in 4 ways and the third can be selected in 3 different ways. Hence, the total number of 3 digit numbers that can be formed are 5 × 4 × 3 = 60
When the pieces of work are mutually exclusive, there are M+N+P ways of doing the complete work.
Number of permutations (or arrangements) of n different things taken all at a time = n!
Number of permutations of n things out of which P1 are alike and are of one type, P2 are alike and are of a second type and P3 are alike and are of a third type and the rest are all different = \( \frac{n!}{P1! * P2! * P3!} \)
Illustration: The number of words formed with the letters of the word Allahabad. Solution: Total number of Letters = 9 of which A occurs four times, L occurs twice and the rest are all different. Total number of words formed = 9! / (4! 2! 1! )
Number of permutations of n different things taken r at a time when repetition is allowed = n × n × n × ... (r times) = n^{r} .
Illustration: In how many ways can 4 rings be worn in the index, ring finger and middle finger if there is no restriction of the number of rings to be worn on any finger? Solution: Each of the 4 rings could be worn in 3 ways either on the index, ring or middle finger. So, four rings could be worn in 3 × 3 × 3 × 3 = 3^{4} ways.
Number of selections of r things out of n identical things = 1 Illustration: In how many ways 5 marbles can be chosen out of 100 identical marbles? Solution: Since, all the 100 marbles are identical Hence, Number of ways to select 5 marbles = 1
Total number of selections of zero or more things out of k identical things = k + 1. This includes the case when zero articles are selected.
Total number of selections of zero or more things out of n different things \(= nC_0 + nC_1 + ... + nC_n = 2^n\)
Corollary: The number of selections of 1 or more things out of n different things = \(= nC_1 + ... + nC_n = 2^n - 1\)
Number of ways of distributing n identical things among r persons when each person may get any number of things = ^{n + r – 1}C_{r–1}
Imagine a situation where 27 marbles have to be distributed amongst 4 people such that each one of them can get any number of marbles (including zero marbles). Then for this situation we have, n = 27(no. of identical objects), r = 4 (no. of people) and the answer of the number of ways this can be achieved is given by: ^{n + r – 1}C_{r–1} = ^{30}C_{3}
Corollary: No. of ways of dividing n non distinct things to r distinct groups are: ^{n–1}C_{r–1} so For non-empty groups only Also, the number of ways in which n distinct things can be distributed to r different persons: = r^{n}
Number of ways of dividing m + n different things in two groups containing m and n things respectively = ^{m + n}C_{n} × ^{m}C_{m} = = (m + n)! /m! n! Or, ^{m+n}C_{m} × ^{n}C_{n} = (m + n)! / n! m!
Number of ways of dividing 2n different things in two groups containing n things = 2n! / n! n! 2!
^{n}C_{r} + ^{n}C_{r – 1} = ^{n + 1}C_{r}
^{n}C_{x} = ^{n}C_{y} so x = y or x + y = n
^{n}C_{r} = ^{n}C_{n – r}
r . ^{n}C_{r} = n . ^{n – 1}C_{r – 1}
^{n}C_{r} / (r + 1) = ^{n + 1}C_{r + 1}/( n + 1)
For ^{n}Cr to be greatest,
if n is even, r = n/2
if n is odd, r = (n + 1)/2 or (n – 1)/2
Number of selections of r things out of n different things
When k particular things are always included = ^{n – k}C_{r} – k
When k particular things are excluded = ^{n – k}C_{r}
When all the k particular things are not together in any selection = ^{n}C_{r} – ^{n – k}C_{r} – k
No. of ways of doing a work with given restriction = total no. of ways of doing it — no. of ways of doing the same work with opposite restriction.
The total number of ways in which 0 to n things can be selected out of n things such that p are of one type, q are of another type and the balance r of different types is given by: (p + 1)(q + 1)(2r– 1).
Total number of ways of taking some or all out of p + q + r things such that p are of one type and q are of another type and r of a third type = (p + 1)(q + 1)(r + 1) – 1 [Only non-empty sets]
\( \frac{nCr}{nCr-1} = \frac{n-r+1}{r} \)
Number of selections of k consecutive things out of n things in a row = n – k + 1
Consider two situations: There are three A, B and C. In the first case, they are arranged linearly and in the other, around a circular table
For the linear arrangement, each arrangement is a totally new way. For circular arrangements, three linear arrangements are represented by one and the same circular arrangement.
So, for six linear arrangements, there correspond only 2 circular arrangements. This happens because there is no concept of a starting point on a circular arrangement. (i.e., the starting point is not defined.) Generalising the whole process, for n!, there corresponds to be (1/n) n! ways.
Number of ways of arranging n people on a circular track (circular arrangement) = (n – 1)!
When clockwise and anti-clockwise observation are not different then number of circular arrangements of n different things = (n – 1)! /2 e.g. the case of a necklace with different beads, the same arrangement when looked at from the opposite side becomes anti-clockwise.
Number of selections of k consecutive things out of n things in a circle = n when k < n = 1 when k = n
Number of terms in \( (a_1 + a_2 + ... + a_n)^m\) is ^{m + n – 1}C_{n–1}
Q. Find the number of terms in (a + b + c)^{2}
n = 3, m = 2
^{m + n – 1}C_{n–1} = ^{4}C_{2} = 6
Corollary: Number of terms in \( [1 + x + x^2 v+ ... + x^n]^m = mn+1 \)
Q. Find the number of zeroes at the end of 1000!
Number of zeroes ending the number represented by n! = [n/5] + [n /5^{2}] + [n/5^{3}] + ... + [n/5^{x}]
[] Shows greatest integer function where 5^{x} ≤ n
[1000/5] + [1000 /5^{2}] + [1000/5^{3}] + ... + [1000/5^{4}]
200 + 40 + 8 + 1 = 249
Q. Find how many exponents of 3 will be there in 24!.
Find how many exponents of 3 is represented by n! = [n/3] + [n /3^{2}] + [n/3^{3}] + ... + [n/3^{x}]
[] Shows greatest integer function where 3^{x} ≤ n
[24/3] + [24/3^{2}] = 8 + 2 = 10
Example : The number of squares and rectangles in the figure are given by:
Number of squares in a square of n × n side = \(1^2 + 2^2 + 3^2 + ... + n^2\)
Number of rectangles in a square of n × n side = \(1^3 + 2^3 + 3^3 + ... + n^3\) . (This includes the number of squares.)
A rectangle having m rows and n columns: The number of squares is given by: m.n + (m – 1)(n – 1) + (m – 2)(n – 2) + ... until any of (m – x) or (n – x) comes to 1
The number of rectangles is given by: (1 + 2 + ... + m)(1 + 2 + ... + n)
Q. The number of squares and rectangles in the following figure are given by:
Number of squares = \(1^2 + 2^2 + 3^2 = 14 \)
Number of rectangles = \( 1^3 + 2^3 + 3^3 = 36 \)
The answer will be given by 6P4
Forming numbers requires an ordered selection. Hence, the answer will be 5P3 .
n For arranging the 7 letters keeping P and Q always together we have to view P and Q as one letter. Let this be denoted by PQ. Then, we have to arrange the letters M, N, O, PQ, R and S in a linear arrangement. Here, it is like arranging 6 letters in 6 places (since 2 letters are counted as one). This can be done in 6! ways. However, the solution is not complete at this point of time since in the count of 6! the internal arrangement between P and Q is neglected. This can be done in 2! ways. Hence, the required answer is 6! × 2!.
5! × 3!
(Answer will be given by the formula: Total number of ways – Number of ways they are always together)
B & S are fixed at the start and the end positions. Hence, we have to arrange E, G, I and N amongst themselves. This can be done in 4! ways.
4! × 2!
There are 4 vowels and 7 consonants in Valedictory. If these vowels have to be kept together, we have to consider AEIO as one letter. Then the problem transforms itself into arranging 8 letters amongst themselves (8! ways). Besides, we have to look at the internal arrangement of the 4 vowels amongst themselves. (4! ways) Hence Answer = 8! × 4!
The significance of at least 5 hats of each kind is that while putting a hat in each box, we have the option of putting either a red or a blue hat. (If this was not given, there would have been an uncertainty in the number of possibilities of putting a hat in a box.) Thus in this question for every task of putting a hat in a box we have the possibility of either putting a red hat or a blue hat. The solution can then be looked at as: there are 5 tasks each of which can be done in 2 ways. Through the MNP rule we have the total number of ways = 2^{5} (Answer).
If we first put 4 Indians around the round table, we can do this in 3! ways. Once the 4 Indians are placed around the round table, we have to place the four Nepalese around the same round table. Now, since the Indians are already placed we can do this in 4! ways (as the starting point is defined when we put the Indians. Try to visualize this around a circle for placing 2 Indians and 2 Nepalese.) Hence, Answer = 3! × 4!
In order to form a number greater than a million we should have a 7 digit number. Since we have only seven digits with us we cannot take 0 in the starting position. View this as 7 positions to fill: _ _ _ _ _ _ _ To solve this question we first assume that the digits are all different. Then the first position can be filled in 6 ways (0 cannot be taken), the second in 6 ways (one of the 6 digits available for the first position was selected. Hence, we have 5 of those 6 digits available. Besides, we also have the zero as an additional digit), the third in 5 ways (6 available for the 2nd position – 1 taken for the second position.) and so on. Mathematically this can be written as: 6 × 6 × 5 × 4 × 3 × 2 × 1 = 6 × 6! This would have been the answer had all the digits been distinct. But in this particular example we have two 2’s and two 3’s which are identical to each other. This complication is resolved as follows to get the answer: (6 * 6)/(2! * 2!)
16C11
When 18 identical white balls are put in a straight line, there will be 19 spaces created. Thus 16 black balls will have 19 places to fill in. This will give an answer of: 19C16 . (Since, the balls are identical the arrangement is not important.)
She will be able to do this as many times as she can form a set of three distinct children from amongst the seven children. This essentially means that the answer is the number of selections of 3 people out of 7 that can be done. Hence, Answer = 7C3
This can be viewed as: The child for whom we are trying to calculate the number of ways is already selected. Then, we have to select 2 more children from amongst the remaining 6 to complete the group. This can be done in 6C2 ways
A distinct sum will be formed by selecting either 1 or 2 or 3 or 4 or 5 or all 6 coins. But from the formula we have the answer to this as : 2^{6} – 1. [Task for the student: How many different sums can be formed with the following coins: 5 rupee, 1 rupee, 50 paisa, 25 paisa, 10 paisa, 3 paisa, 2 paisa and 1 paisa? Hint: You will have to subtract some values for double counted sums.]
Since the 3 persons are entering during the journey they could have entered at the: 1st station (from where they could have bought tickets for the 2nd, 3rd, 4th or 5th stations or for Pune total of 5 tickets.) 2nd station (from where they could have bought tickets for the 3rd, 4th or 5th stations or for Pune Æ total of 4 tickets.) 3rd station (from where they could have bought tickets for the 4th or 5th stations or for Pune Æ total of 3 tickets.) 4th station (from where they could have bought tickets for the 5th station or for Pune Æ total of 2 tickets.) 5th station (from where they could have bought a ticket for Pune Æ total of 1 ticket.) Thus, we can see that there are a total of 5 + 4 + 3 + 2 + 1 = 15 tickets available out of which 3 tickets were selected. This can be done in 15C3 ways (Answer).
A decagon has 10 vertices. A line is formed by selecting any two of the ten vertices. This can be done in 10C2 ways. However, these 10C2 lines also count the sides of the decagon. Thus, the number of diagonals in a decagon is given by: 10C2 – 10 (Answer) Triangles are formed by selecting any three of the ten vertices of the decagon. This can be done in 10C3 ways (Answer).
If all 18 points were non-collinear then the answer would have been 18C2 . However, in this case 18C2 has double counting since the 5 collinear points are also amongst the 18. These would have been counted as 5C2 whereas they should have been counted as 1. Thus, to remove the double counting and get the correct answer we need to adjust by reducing the count by (5C2 – 1). Hence, Answer =18C2 – (5C2 – 1) = 18C2 – 5C2 + 1
The triangles will be given by 18C3 – 5C3.
2^{10} = 1024 unique sequences are possible. Option (d) is correct
For n people there are always nC2 shake hands. Thus, for 10 people shaking hands with each other the number of ways would be 10C2 = 45.
If the children are A, B, C, D we have to consider A & B as one child. This, would give us 3! ways of arranging AB, C and D. However, for every arrangement with AB, there would be a parallel arrangement with BA. Thus, the correct answer would be 3! × 2! = 12 ways. Option (b) is correct
There would be 11C2 combinations of 2 people taken 2 at a time for comparison. 11C2 = 55
2^{4} – 1 = 15 sums of money can be formed. Option (b) is correct
2^{4} – 1 = 15. Hence, option (b) is correct.
Numbers starting with 12 – 7 numbers Numbers starting with 13 – 6 numbers; 14 – 5, 15 – 4, 16 – 3, 17 – 2, 18 – 1. Thus total number of numbers starting from 1 is given by the sum of 1 to 7 = 28. Number of numbers starting from 2- would be given by the sum of 1 to 6 = 21 Number of numbers starting from 3- sum of 1 to 5 = 15 Number of numbers starting from 4 – sum of 1 to 4 = 10 Number of numbers starting from 5 – sum of 1 to 3 = 6 Number of numbers starting from 6 = 1 + 2 = 3 Number of numbers starting from 7 = 1 Thus a total of: 28 + 21 + 15 + 10 + 6 + 3 + 1 = 84 such numbers. Option (d) is correct
The given condition can get achieved if we were to use 17 boys and 13 girls. In such a case both statement I and II are correct. Hence, option (c) is correct.
The placement of balls can be 3, 1, 1 and 2, 2, 1. For 3, 1, 1- If we place 3 balls in the top row, there would be 3C1 ways of choosing a place for the ball in the second row and 3C1 ways of choosing a place for the ball in the third row. Thus, 3C1 × 3C1 = 9 ways. Similarly there would be 9 ways each if we were to place 3 balls in the second row and 3 balls in the third row. Thus, with the 3, 1, 1 distribution of 5 balls we would get 9 + 9 + 9 = 27 ways of placing the balls. We now need to look at the 2, 2, 1 arrangement of balls. If we place 1 ball in the first row, we would need to place 2 balls each in the second and the third rows. In such a case, the number of ways of arranging the balls would be 3C1 × 3C2 × 3C2 = 27 ways. (choosing 1 place out of 3 in the first row, 2 places out of 3 in the second row and 2 places out of 3 in the third row). Similarly if we were to place 1 ball in the second row and 2 balls each in the first and third rows we would get 27 ways of placing the balls and another 27 ways of placing the balls if we place 1 ball in the third row and 2 balls each in the other two rows. Thus with a 2, 2, 1 distribution of the 5 balls we would get 27 + 27 + 27 = 81 ways of placing the balls. Hence, total number of ways = Number of ways of placing the balls with a 3,1,1 distribution of balls + number of ways of placing the balls with a 2, 2, 1 distribution of balls = 27 + 81 = 108. Hence, option (d) is correct.
If 5 letters go into the correct envelopes the sixth would automatically go into it’s correct envelope. Thus, there is no possibility when exactly 5 letters are correct and 1 is wrong. Hence, option (a) is correct.
In the first cell, we have 3 options of placing a ball. Suppose we were to place a red ball in the first cell- then the second cell can only be filled with either black or white – so 2 ways. Subsequently there would be 2 ways each of filling each of the cells (because we cannot put the colour we have already used in the previous cell). Thus, the required number of ways would be 3 × 2 × 2 × 2 = 24 ways. Hence, option (c) is correct.
Look for the smallest triangles first—there are 12 of them. Then, look for the triangles which are equal to half the rectangle—there are 12 of them. Besides, there are 4 bigger triangles (spanning across 2 rectangles). Thus a total of 28 triangles can be seen in the figure. Hence, option (a) is correct
If the students are A, B, C, D, E and F- we can have 6C3 groups in all. However, if we have to count groups in which a particular student (say A) is always selected- we would get 5C2 = 10 ways of doing it. Hence, option (c) is correct
All 3 dice have twos – 1 case. Two dice have twos: This can principally occur in 3 ways which can be broken into: If the first two dice have 2- the third dice can have 1, 3, 4, 5 or 6 = 5 ways. Similarly, if the first and third dice have 2, the second dice can have 5 outcomes Æ 5 ways and if the second and third dice have a 2, there would be another 5 ways. Thus a total of 15 outcomes if 2 dice have a 2. With only 1 dice having a two- If the first dice has 2, the other two can have 5 × 5 = 25 outcomes. Similarly 25 outcomes if the second dice has 2 and 25 outcomes if the third dice has 2. A total of 75 outcomes. Thus, a total of 1 + 15 + 75 = 91 possible outcomes with at least. Hence, option (c) is correct.
All words staring with A, C, H, I and N would be before words starting with S. So we would have 5! Words (= 120 words) each starting with A, C, H, I and N. Thus, a total of 600 words would get completed before we start off with S. SACHIN would be the first word starting with S, because A, C, H, I, N in that order is the correct alphabetical sequence. Hence, Sachin would be the 601 st word. Hence, option (c) is correct.
The arrangements can be [3 & 1 & 1 or 1 & 3 & 1 or 1 & 1 & 3] or 2 & 2 & 1 or 2 & 1 & 2 or 1 & 2 and 2. Total number of ways = 3 × 5C3 × 2C1 × 1C1 + 3 × 5C2 × 3C2 × 1C1 = 60 + 90 = 150 ways Hence, option (c) is correct.
n The selection can be done in the following ways: 2 boys from Amit’s friends and 2 girls from his wife’s friends OR 1 boy & 1 girl from Amit’s friends and 1 boy and 1 girl from his wife’s friends OR 2 girls from Amit’s friends and 2 boys from his wife’s friends. The number of ways would be: 2C2 × 2C2 + 3C1 × 2C1 × 3C1 × 2C1 + 3C2 × 3C2 = 1 + 36 + 9 = 46 ways.
The ways of placing the balls would be 5, 1, 1, 1 (4!/3! = 4 ways); 4, 2, 1 & 1 (4!/2! = 12 ways); 3, 3, 1, 1 (4!/2! × 2! = 6 ways); 3, 2, 2, 1 (4!/2! = 12 ways) and 2, 2, 2, 2 (1 way). Total number of ways = 4 + 12 + 6 + 12 + 1 = 35 ways. Hence, option (b) is correct.
The number of squares would be 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} = 91. Hence, option (c) is correct
All groups – groups with C and D together = 5C3 – 3C1 = 10 – 3 = 7
The thought process for this question would be: All arrangements (12C3) – Arrangements where all 3 balls are in the same row (3 × 4C3) – arrangements where all 3 balls are in the same straight line diagonally (4 arrangements) = 12C3 – 3 × 4C3 – 4 = 220 – 12 – 4 = 204 ways. Hence, option (c) is correct
All numbers – numbers having no numbers repeated = 1000 – 9 × 8 × 7 = 1000 – 504 = 496 numbers. Hence, option (c) is correct
4C2 × 2! × 6C3 × 3! = 6 × 2 × 20 × 6 = 1440. Hence, option (c) is correct
Let C1, C2, C3, C4 and C5 be the 5 distinct colours which have no repetition. For being definitely sure that we have picked up 2 balls of the same colour we need to consider the worst case situation. In the above distribution of balls each set has exactly 1 ball which is unique in it’s colour while the colours of the other two balls are shared at least once in one of the other sets. In such a case, the worst scenario would be if we pick up the first 10 balls and they all turn out to be of different colours. The 11 th ball has to be of a colour which has already been taken. Thus,if we were to pick out 11 balls we would be sure of having at least 2 balls of the same colour. Hence, option (d) is correct.
5^{4} would be the total number of ways in which the questions can be answered. Out of these there would be only 1 way of getting all 4 correct. Thus, there would be 624 ways of not getting all answers correct.
The 6 balls must be on either of the middle rows. This can be done in 2 ways. Once, we put the 6 balls in their single horizontal row- it becomes evident that for placing the 2 remaining balls on a straight line there are 2 principal options: 1. Placing the two balls in one of the four rows with two squares. In this case the number of ways of placing the balls in any particular row would be 1 way (since once you were to choose one of the 4 rows, the balls would automatically get placed as there are only two squares in each row.) Thus the total number of ways would be 2 × 4 × 1 = 8 ways. 2. Placing the two balls in the other row with six squares. In this case the number of ways of placing the 2 balls in that row would be 6C2 . This would give us 2C1 × 1 × 6C2 = 30 ways. Total is 30 + 8 = 38 ways. Hence, option (a) is correct
The number of players at the start of the tournament cannot be 8, 10 or 12 because in each of these cases the total number of matches would be less than 75 (as 8C2,10C2 and12C2 are all less than 75.) This only leaves 15 participants in the tournament as the only possibility. Hence, option (d) is correct.
The maximum triangles would be in case all these 6 points are non-collinear. In such a case the number of triangles is 6C3 = 20. Statement I is incorrect. Statement II is correct because if we take the position that A and B coincide on the first line, C & D coincide on the second line, E & F coincide on the third line and all these coincidences happen at 3 points which are on the same straight line- in such a case there would be 0 triangles formed. Hence, option (b) is correct
First select the two men. This can be done in 4C2 ways. Let us say the men are A, B, C and D and their respective wives are a, b, c and d. If we select A and B as the two men then while selecting the women there would be two cases as seen below: Total number of ways of doing this = 4C2 × 2 × 2C1 = 24 ways. Hence, the required answer is 18 + 24 = 42 ways. Hence, option (d) is correct.
The number of numbers formed would be given by 5 × 4 × 3 (given that the first digit can be filled in 5 ways, the second in 4 ways and the third in 3 ways – MNP rule).
The first digit can only be 2 (1 way), the second digit can be filled in 7 ways, the third in 6 ways and the fourth in 5 ways. A total of 1 × 7 × 6 × 5 = 210 ways.
Each invitation card can be sent in 4 ways. Thus, 4 × 4 × 4 × 4 × 4 × 4 = 4^{6}
In this case since nothing is mentioned about whether the prizes are identical or distinct we can take the prizes to be distinct (the most logical thought given the situation). Thus, each prize can be given in 8 ways — thus a total of 8^{5} ways.
We need to assume that the 7 Indians are 1 person, so also for the 6 Dutch and the 5 Pakistanis. These 3 groups of people can be arranged amongst themselves in 3! ways. Also, within themselves the 7 Indians the 6 Dutch and the 5 Pakistanis can be arranged in 7!, 6! And 5! ways respectively. Thus, the answer is 3! × 7! × 6! × 5!.
Use the MNP rule to get the answer as 5 × 4 = 20.
An IITian can make it to IIMs in 2 ways, while a CA can make it through in 3 ways. Required ratio is 2:3.
For a straight line we just need to select 2 points out of the 8 points available. 8C2 would be the number of ways of doing this.
Use the property nCr = nCn-r to see that the two values would be equal at n = 11 since 11C3 = 11C8
Solve this one through options. If you pick up option (a) it gives you 12 participants in the tournament. This means that there are 10 men and 2 women. In this case there would be 2 × 10C2 = 90 matches amongst the men and 2 × 10C1 × 2C1 = 40 matches between 1 man and 1 woman. The difference between number of matches where both participants are men and the number of matches where 1 participant is a man and one is a woman is 90 – 40 = 50 – which is not what is given in the problem. With 15 participants Æ 11 men and 2 women. In this case there would be 2 × 11C2 = 110 matches amongst the men and 2 × 11C1 × 2C1 = 44 matches between 1 man and 1 woman. The difference between number of matches where both participants are men and the number of matches where 1 participant is a man and one is a woman is 110 – 44 = 66 – which is the required value as given in the problem. Thus, option (b) is correct
Based on the above thinking we get that since there are 15 players and each player plays each of the others twice, the number of games would be 2 × 15C2 = 2 × 105 = 210
Number of even numbers = 6 × 6 × 6 × 3 = 648
For each selection there are 3 ways of doing it. Thus, there are a total of 3 × 3 × 3 × 3 × 3 = 243.
One digit no. = 5; Two digit nos = 5 × 4 = 20; Three digit no = 5 × 4 × 3 = 60; four digit no = 5 × 4 × 3 × 2 = 120; Five digit no.= 5 × 4 × 3 × 2 × 1 = 120 Total number of nos = 325
With one green ball there would be six ways of doing this. With 2 green balls 5 ways, with 3 green balls 4 ways, with 4 green balls 3 ways, with 5 green balls 2 ways and with 6 green balls 1 way. So a total of 1 + 2 + 3 + 4 + 5 + 6 = 21 ways
Each of the first, third and fourth options can be obviously seen to be true— no mathematics needed there. Only the second option can never be true. In order to think about this mathematically and numerically— think of a party of 3 persons say A, B and C. In order for the second condition to be possible, each person must know a different number of persons. In a party with 3 persons this is possible only if the numbers are 0, 1 and 2. If A knows both B and C (2), B and C both would know at least 1 person— hence it would not be possible to create the person knowing 0 people. The same can be verified with a group of 4 persons i.e., the minute you were to make 1 person know 3 persons it would not be possible for anyone in the group to know 0 persons and hence you would not be able to meet the condition that every person knows a different number of persons
Since there are 11 symmetric letters, the number of passwords that can be formed would be 11 × 10 × 9 × 8 = 7920.
This would be given by the number of passwords having: 1 symmetric and 2 asymmetric letters + 2 symmetric and 1 asymmetric letter + 3 symmetric and 0 asymmetric letters 11C1 × 15C2 × 3! + 11C2 × 15C1 × 3! + 11C3 × 3! = 11 × 105 × 6 + 55 × 15 × 6 + 11 × 10 × 9 = 6930 + 4950 + 990 = 12870
The white square can be selected in 32 ways and once the white square is selected 8 black squares become ineligible for selection. Hence, the black square can be selected in 24 ways. 32 × 24 = 768.
A million is 1000000 (i.e. the first seven digit number). So we need to find how many numbers of less than 7 digits can be formed using the digits 0,7 and 8. Number of 1 digit numbers = 2 Number of 2 digit numbers = 2 × 3 = 6 Number of 3 digit numbers = 2 × 3 × 3 = 18 Number of 4 digit numbers = 2 × 3 × 3 × 3 = 54 Number of 5 digit numbers = 2 × 3 × 3 × 3 × 3 = 162 Number of 6 digit numbers = 2 × 3 × 3 × 3 × 3 × 3 = 486 Total number of numbers = 728
If the number of teams is n, then nC2 should be equal to 45. Trial and error gives us the value of n as 10.
From 5 bananas we have 6 choices available (0, 1, 2, 3, 4 or 5). Similarly 4 mangoes and 4 almonds can be chosen in 5 ways each. So total ways = 6 × 5 × 5 = 150 possible selections. But in this 150, there is one selection where no fruit is chosen. So required no. of ways = 150 – 1 = 149
For each book we have two options, give or not give. Thus, we have a total of 2 14 ways in which the 14 books can be decided upon. Out of this, there would be 1 way in which no book would be given. Thus, the number of ways is 2^{14} – 1
The number of ways in which at least 1 Archer book is given is (2^{5}–1). Similarly, for Sheldon and Grisham we have (2^{3}–1) and (2^{6}–1). Thus required answer would be the multiplication of the three (2^{5}–1)(2^{3}–1)(2^{6}–1)
For each question we have 3 choices of answering the question (2 internal choices + 1 nonattempt). Thus, there are a total of 3^{15} ways of answering the question paper. Out of this there is exactly one way in which the student does not answer any question. Thus there are a total of 3^{15} – 1 ways in which at least one question is answered
The digits are 1, 6, 7, 8, 7, 6, 1. In this seven-digit no. there are four add places and three even places— OEOEOEO. The four odd digits 1, 7, 7, 1 can be arranged in four odd places in [4!/2! × 2] = 6 ways [as 1 and 7 are both occurring twice]. The even digits 6, 8, 6 can be arranged in three even places in 3!/2! = 3 ways. Total no. of ways = 6 × 3 = 18
We have no girls together, let us first arrange the 5 boys and after that we can arrange the girls in the spaces between the boys. Number of ways of arranging the boys around a circle = [5 – 1]! = 24. Number of ways of arranging the girls would be by placing them in the 5 spaces that are formed between the boys. This can be done in 5P3 ways = 60 ways. Total arrangements = 24 × 60 = 1440
Books of interest = 7, books to be borrowed = 3 Case 1— Quants book is taken. Then D.I book can also be taken. So Amita is to take 2 more books out of 6 which she can do in 6C2 = 15 ways. Case 2— If Quants book has not been taken, the D.I book would also not be taken. So Amita will take three books out of 5 books. This can be done in 5C3 = 10 ways. So total ways = 15 + 10 = 25 ways.
We have to select 5 out of 12. If Radha and Mohan join- then we have to select only 5 – 2 = 3 dancers out of 12 – 2 = 10 which can be done in 10C3 = 120 ways. If Radha and Mohan do not join, then we have to select 5 out of 12 – 2 = 10-> 10C5 = 252 ways. Total number of ways = 120 + 252 = 372.
The unit digit can either be 2, 3, 4, 5 or 6. When the unit digit is 2, the number would be even and hence will be divisible by 2. Hence all numbers with unit digit 2 will be included which is equal to 5! Or 120. When the unit digit is 3, then in every case the sum of the digits of the number would be 21 which is a multiple of 3. Hence all numbers with unit digit 3 will be divisible by 3 and hence will be included. Total number of such numbers is 5! or 120. Similarly for unit digit 5 and 6, the number of required numbers is 120 each. When the unit digit is 4, then the number would be divisible by 4 only if the ten’s digit is 2 or 6. Total number of such numbers is 2 × 4! or 48. Hence total number of required numbers is (4 × 120) + 48 = 528.
As we need to find the maximum number of trials, so we have to assume that the required ball in every box is picked as late as possible. So in the third box, first two balls will be red and black. Hence third trial will give him the required ball. Similarly, in fourth box, he will get the required ball in fourth trial and in the fifth box, he will get the required ball in fifth trial. Hence maximum total number of trials required is 3 + 4 + 5 = 12.
Since every player needs to win only 1 match to move to the next round, therefore the 1st round would have 32 matches between 64 players out of which 32 will be knocked out of the tournament and 32 will be moved to the next round. Similarly in 2nd round 16 matches will be played, in the 3rd round 8 matches will be played, in 4th round 4 matches, in 5th round 2 matches and the 6th round will be the final match. Hence total number of rounds will be 6 (2^{6} = 64).
Total number of pairs of men that can be selected if the adjacent ones are also selected is NC2. Total number of pairs of men selected if only the adjacent ones are selected is N. Hence total number of pairs of men selected if the adjacent ones are not selected is NC2 – N. Since the total time taken is 88 min, hence the number of pairs is 44. Hence, NC2 – N = 44 so N = 11
Q. Number of ways of arranging MALAYALAM such that vowels are never together.
A. Total ways = 5 letters and 4 vowels; Vowels can be treated as single entity giving us 6 alphabets; So 6P6 = 6! but M and L are repeated twice so removing duplicate entries we get 6!/(2!*2!) ways.
Total ways in which 9 letters can be arranged are = 9!/(2!*2!*4!) ways removing duplicates for 2 M's, 2 L's and 4 A's
Total ways vowels are never together = Total ways - Total ways were they are together
Q. How many ways can a cricket team of 11 be chosen out of batch of 15 players.
A. ways = 15C11 = 15C4
Q. How many ways to select a committee of 3 men and 2 ladies from 6 men and 5 ladies.
A. Men can be chosen in 6C3 ways and women in 5C2 ways.
Total ways are = 6C3 * 5C2.
Q. There are 6 boxes numbered 1,2,… 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
5
21
33
60
Ans . B
GRRRRR, RGRRRR, RRGRRR, RRRGRR, RRRRGR, RRRRRG GGRRRR, RGGRRR, RRGGRR, RRRGGR, RRRRGG GGGRRR, RGGGRR, RRGGGR, RRRGGG GGGGRR, RGGGGR, RRGGGG GGGGGR, RGGGGG GGGGGG
Hence 21 ways.
Q. Let T be the set of integers {3, 11, 19, 27, …, 451, 459, 467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is
32
28
29
30
Ans . D
T_{n} = a + (n-1)d
467 = 3 + (n – 1)8 ⇒ n = 59
Half of n = 29 terms 29th term is 227 and 30 th term is 235 and when these two terms are added the sum is less than 470. Hence the maximum possible values the set S can have is 30.
Q. The number of positive integers n in the range 12 ≤ n ≤ 40 such that the product (n - 1)(n - 2).. 3.2.1 is not divisible by n is
5
7
13
14
Ans . B
From 12 to 40, there are 7 prime numbers, i.e., 13, 17, 19, 23, 29, 31 and 37 such that (n – 1)! is not divisible by any of them.
Q. A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is a graph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any point through a sequence of edges. The number of edges, e, in the graph must satisfy the condition
11 ≤ e ≤ 66
10 ≤ e ≤ 66
11 ≤ e ≤ 65
0 ≤ e ≤ 11
Ans . A
The least number of edges will be when one point is connected to each of the other 11 points, giving a total of 11 lines. One can move from any point to any other point via the common point. The maximum edges will be when a line exists between any two points. Two points can be selected from 12 points in
^{12}C_{2}i.e. 66 lines
Q. In a certain examination paper, there are n questions. For j = 1,2 …n, there are 2^{n-j} students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is
12
11
10
9
Ans . A
Let us say there are only 3 questions. Then, there are
2^{3-1} = 4 students who have done 1 or more questions wrongly, 2^{3-2} = 2 students who have done 2 or more questions wrongly and 2^{3-3} = 1 student who must have done all 3 wrongly. Thus, total number of wrong answers = 4 + 2 + 1 = 7 = 2^{3} – 1 = 2^{n} – 1.
= 4095 = 2^{12} – 1. Thus n = 12
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