Aptitude :: Permutation And CombinationPage 1
1.  From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?  (a)256  (b)520  (c)756  (d)840 

Answer is: CWe may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways = (^{7}C_{3} x ^{6}C_{2}) + (^{7}C_{4} x ^{6}C_{1}) + (^{7}C_{5}).
Required number of ways = [(7 x 6 x 5 )/(3 x 2 x 1)] x [(6 x 5)/(2 x 1)] + (^{7}C_{3} x ^{6}C_{1}) + (^{7}C_{2}).
Required number of ways = 525 + [(7 x 6 x 5)/(3 x 2 x 1)] x 6 + [(7 x 6)/(2 x 1)].
Required number of ways = 525 + 210 + 21 = 756
2.  In how many different ways can the letters of the word LEADING be arranged in such a way that the vowels always come together?  (a)480  (b)512  (c)720  (d)840 

Answer is: CThe word LEADING has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
3.  In how many different ways can the letters of the word CORPORATION be arranged so that the vowels always come together?  (a)4500  (b)50400  (c)65010  (d)72000 

Answer is: BIn the word CORPORATION, we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = 7!/2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5!/3! = 20 ways.
Required number of ways = (2520 x 20) = 50400.
4.  Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?  (a)24500  (b)25200  (c)27300  (d)28100 

Answer is: BNumber of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (^{7}C_{3} x ^{4}C_{2}) = [(7 x 6 x 5)/(3 x 2 x 1)] x [(4 x 3)/(2 x 1) = 210
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves = 5!
Number of ways of arranging 5 letters among themselves = 5 x 4 x 3 x 2 x 1
Number of ways of arranging 5 letters among themselves = 120.
Required number of ways = (210 x 120) = 25200.
5.  In how many ways can the letters of the word LEADER be arranged?  (a)120  (b)260  (c)320  (d)360 

Answer is: DThe word LEADER contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
Required number of ways = 6!/((1!)(2!)(1!)(1!)(1!)) = 360.
6.  In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?  (a)200  (b)209  (c)305  (d)309 

Answer is: BWe may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways = (^{6}C_{1} x ^{4}C_{3}) + (^{6}C_{2} x ^{4}C_{2}) + (^{6}C_{3} x ^{4}C_{1}) + (^{6}C_{4})
Required number of ways = (^{6}C_{1} x ^{4}C_{1}) + (^{6}C_{2} x ^{4}C_{2}) + (^{6}C_{3} x ^{4}C_{1}) + (^{6}C_{2})
Required number of ways = (6 x 4) + [(6 x 5)/(2 x 1)] x [(4 x 3)/(2 x 1)] + [(6 x 5 x 4)/(3 x 2 x 1)] x [4] + [(6 x 5)/(2 x 1)]
Required number of ways = (24 + 90 + 80 + 15)
Required number of ways = 209
7.  How many 3digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?  (a)15  (b)20  (c)25  (d)30 

Answer is: BSince each desired number is divisible by 5, so we must have 5 at the unit place. So, there is one way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
8.  How many words can be formed by using all letters of the word BIHAR ?  (a)120  (b)60  (c)180  (d)260 

Answer is: AThe word BIHAR contains 5 different letters.
Required number of words = 5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120
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