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 |
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Answer is: CWe may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5).
Required number of ways = [(7 x 6 x 5 )/(3 x 2 x 1)] x [(6 x 5)/(2 x 1)] + (7C3 x 6C1) + (7C2).
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 |
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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 |
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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 |
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Answer is: BNumber of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3 x 4C2) = [(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 |
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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 |
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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 = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
Required number of ways = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
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 3-digit 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 |
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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 |
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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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