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# In how many ways can a committee of 4 persons be formed out of 8 people

The term â€ścombinationâ€ť is thrown around loosely, and usually in the wrong way. Things like, â€śHey, whatâ€™s the suitcase lock combination?â€ť are said But what one really ought to be saying is â€śHey, whatâ€™s the suitcase lock permutation?â€ť So whatâ€™s the difference? And what exactly are a permutation and combination? They are two very different terms. Let’s learn about them in detail,

### Permutation

A permutation is an act of arranging objects or given quantity maybe numbers from a group of objects or collection given in a particular order as per given conditions. Example – How many 2 letter words are there that can be formed by using the letters in the word LATE? Answer is 4P2 (pronounced as 4 p 2) = 4!/(4 – 2)! = 4!/(2)! = (4 Ă— 3 Ă— 2 Ă— 1)/(2 Ă— 1) = 24/2 = 12.

### Combination

The combination is the way of selecting the objects or given quantity maybe numbers from a group of objects or collection from a group of objects or collection, in such a way that the order of the objects does not matter. For example – how many groups of 2 people can be selected from 4 people? Answer – 4 C 2(pronounced as 4 C 2) = 4!/(4 – 2)!(2)! =  4!/(2)!(2)! = (4 Ă— 3 Ă— 2 Ă— 1)/(2 Ă— 1)(2 Ă— 1) = 24/4 = 6.

The formula for permutations and combinations

• The formula for permutations is: nPr = n!/(n – r)!
• The formula for combinations is: nCr = n!/[r! (n – r)!]

### Find the number of committees of the size of 4 formed from 8 people.

Here, analyzing the problem given, the hint is to select 4 people from 8 people. The biggest confusion is whether to apply permutation or combination? So start thinking way that whether the order of people will make a difference? if yes then go for permutation otherwise it’s a combination that will solve the question.

In this question let’s first select p1, p2, p3, p4 in the written order and then select p2, p1, p4, p3. Does that make a different group here? The answer is no. Because irrespective of order the first, second, third, fourth person have been selected. So here, go with the combination as the order doesn’t matter.

Applying combination using formula, nCr = n!/[r! (n – r)!]

Here, n = 8 and r = 4

So 8C4 = 8!/(8 – 4)!(4)! = 8!/(4)!(4)! = (8 Ă— 7 Ă— 6 Ă— 5 Ă— 4 Ă— 3 Ă— 2 Ă— 1) / (4 Ă— 3 Ă— 2 Ă— 1)(4 Ă— 3 Ă— 2 Ă— 1) = 70

### Similar Problems

Question 1: Find the number of committees of the size of 3 formed from 5 people.

5C3 =  5!/(5 – 3)!(3)! = 5!/(2)!(3)! = (5 Ă— 4 Ă— 3 Ă— 2 Ă— 1) / (2 Ă— 1)(3 Ă— 2 Ă— 1) = 10

Question 2: How many different committees of 3 members can be chosen out of 5 people in a group so that one particular person is always chosen?

Since one particular person is always to be taken from the available 5 people in the committee of the 3. So in fact, choose 2 persons from the remaining 4 and that can be done in C(4, 2) = 4C2 = 4!/2! 2! = 6 number of ways.

Question 3: How many committees of 5 consisting of 3 men and 2 women can be formed from 8 men and 6 women?

Well, one can form 8 choose 3 groups of men, and for each of those, one can choose any of the 6 choose 2 groups of women.

nCr = n!/[r! (n – r)!]

= 6C2 = (6)!/((2!)(6 – 2)!) = 15

= 8C3 = (8)!/((3!)(8 – 3)!) = 56

So, 56 Ă— 15 = 840 possible combinations, assuming one doesnâ€™t care about anything other than the number of men, the number of women.

If one of those people is to be the chair, then there are for each possible group, 5 possible chairs, so multiply that by 5, 840 Ă— 5 = 4200

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