PERMUTATIONS AND COMBINATIONS

Ans .

870

1. Explanation :

We have, 30!/28! = 30x29x(28!)/28! = (30x29) = 870.

Ans .

205320,24

1. Explanation :

(i) $$_{60}^{3}\textrm{P}$$ = 60!/(60-3)! = 60!/57! = 60x59x58x(57!)/57! = (60x59x58) = 205320.

(ii) $$_{4}^{4}\textrm{P}$$ = 4! = (4x3x2x1) = 24.

Ans .

120,4950,1

1. Explanation :

(i) $$_{8}^{10}\textrm{C}$$ = $$\frac{10*9*8}{3!}$$ = 120.

(ii) $$_{98}^{100}\textrm{C}$$=$$_{(100-98)}^{100}\textrm{C}$$ = $$\frac{100*99}{2!}$$ = 4950.

(iii) $$_{50}^{50}\textrm{C}$$= 1. [$$_{n}^{n}\textrm{C}$$ = 1]

Ans .

120

1. Explanation :

The word BIHAR contains 5 different letters.

Required number of words = $$_{5}^{5}\textrm{P}$$ = 5! = (5x4x3x2x1) = 120.

Ans .

4320

1. Explanation :

Given word contains 8 different letters. When the vowels AUE are always together,

we may suppose them to form an entity, treated as one letter.

Then, the letters to be arranged are DGNTR (AUE).

Then 6 letters to be arranged in $$_{6}^{6}\textrm{P}$$ = 6! = 720 ways.

The vowels in the group (AUE) may be arranged in 3! = 6 ways.

Required number of words = (720x6) = 4320.

Ans .

72

1. Explanation :

The given word contains 5 different letters.

Taking the vowels EA together, we treat them as one letter.

Then, the letters to be arranged are XTR (EA).

These letters can be arranged in 4! = 24 ways.

The vowels EA may be arranged amongst themselves in 2! = 2 ways.

Number of words, each having vowels together = (24x2) = 48 ways.

Total number of words formed by using all the letters of the given words

= 5! = (5x4x3x2x1) = 120.

Number of words, each having vowels never together = (120-48) = 72.

Ans .

2160

1. Explanation :

In the given word, we treat the vowels IEO as one letter.

Thus, we have DRCTR (IEO).

This group has 6 letters of which R occurs 2 times and others are different.

Number of ways of arranging these letters = 6!/2! = 360.

Now 3 vowels can be arranged among themselves in 3! = 6 ways.

Required number of ways = (360x6) = 2160.

Ans .

1365

1. Explanation :

Required number of ways = $$_{11}^{15}\textrm{C}$$ = $$_{(15-11)}^{15}\textrm{C}$$ = $$_{4}^{11}\textrm{C}$$

= $$\frac{15*14*13*12}{4*3*2*1}$$ = 1365.

Ans .

200

1. Explanation :

(3 men out 6) and (2 ladies out of 5) are to be chosen.

Required number of ways = ($$_{3}^{6}\textrm{C}$$x$$_{2}^{5}\textrm{C}$$) = [$$\frac{6*5*4}{3*2*1}$$] x [$$\frac{5*4}{2*1}$$] = 200