Learn and practice Problems on Permutation and Combination with easy explaination and shortcut tricks. All questions and answers on Permutation and Combination covered for various Competitive Exams.
Problems on Permutation and Combination:
1) In what ways the letters of the word “RUMOUR” can be arranged?
The word PUZZLE has 6 different letters, but ATQ, the vowels should always come together. Now, let the vowels UE as a single entity. Therefore, the number of letters is PZZL = 4, and UE = 1 Since the total number of letters = 4+1 = 5 So the arrangement would be in 5P5 = = = 5! = 5*4*3*2*1 = 120 ways.
Note: we know that 0! = 1
Now, the vowels AE can arrange in 2 different ways, i.e., 2P2 = 2! = 2*1 = 2 ways
Hence, the new words that can be formed = 120 *2 = 240.
3) In what ways can a group of 6 boys and 2 girls be made out of the total of 7 boys and 3 girls?
The combination of 6 boys out of 7 and 2 girls out of 3 can be represented as 7C6 + 3C2 Therefore, the required number of ways = 7C6 * 3C2 = 7C(7-6) * 3C(3-2) = = 21
Hence, in 21 ways the group of 6 boys and 2 girls can be made.
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4) Out of a group of 7 boys and 6 girls, five boys are selected to form a team so that at least 3 boys are there on the team. In how many ways can it be done?
We may have 5 men only, 4 men and 1 woman, and 3 men and 2 women in the committee.
So, the combination will be
as we know that
nCr=
So, (7C3 * 6C2) + (7C4 * 6C1) + (7C5) Or, + +
Or, 525 +210+21 = 756
So, there are 756 ways to form a committee.
5) A box contains 2 red balls, 3 black balls, and 4 white balls. Find the number of ways by which 3 balls can be drawn from the box in which at least 1 black ball should be present.
A number is divisible by 5 if the number ends with 0 or 5, but we don’t have 0 in the given digits that means 5 should come at the unit place.
Now, one of the remaining 5 digits (2, 3, 4, 6, and 7) can come at the tens place. Similarly, we can fill the hundreds place by one of the remaining 4 digits. Therefore, the thousands place can be filled by one of the remaining 3 digits. Hence, the required number of the numbers = 1*5*4*3 = 60.
8) In what ways the letters of the word “CRICKET” can be arranged to form the different new words so that the vowels always come together?
The word CRICKET has 7 different letters, but ATQ, the vowels should always come together. Now, let the vowels IE as a single entity. Therefore, the number of letters is CRCKT = 5 in which C is repeated twice, and IE = 1 Since the total number of letters = 5+1 = 6 So the arrangement (permutation) would be = = = 5*4*3= 60 ways.
Note: Here 2! is taken in the denominator, because the letter C is repeated twice.
Now, the vowels IE can be arranged in 2 different ways, i.e., 2P2 = 2! = 2*1 = 2 ways.
So, the new words that can be formed = 60*2 = 120.
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9) In what ways the letters of the word ?MACHINE? can be arranged so that the vowels occupy only the odd positions?
The word machine consists of 7 letters in which there are 3 vowels and 4 consonants. ATQ, the vowels A, I, E can be placed at any of the position out of 1, 3, 5, and 7. That means the number of ways to arrange the vowels = 4P3 = = = 4*3*2*1 = 24
Similarly, the number of ways to arrange the consonants = 4P4 = = 4*3*2*1 = 24
Note: we know that 0! = 1
Therefore the required numbers of ways = 24*24 = 576
10) In what ways the letters of the word ?ACTORS? can arrange so that the vowels occupy only the even positions?
The word ACTORS consist of 6 letters in which there are 2 vowels and 4 consonants ATQ, the vowels A, O can be placed at any of the position out of 2, 4, and 6. That means the number of ways to arrange the vowels = 3P2 = = = 3*2 = 6 ways
Similarly, the number of ways to arrange the consonants = 4P4 = = 4*3*2*1 = 24 ways
Note: we know that 0! = 1
Therefore the required numbers of ways = 6*24 = 144