A combination is a sample (group, set, committee, collection) of items drawn fro

Question

A combination is a sample (group, set, committee, collection) of items drawn from a sample space where order doesn't distinguish one sample of things from another sample of the same things. For example, a poker hand formed by drawing cards 8H, 3C, KD, OS, and 2D is the same hand formed by drawing OS, 3C, KD, 8H, and 2D. Suppose the sample space contains things a, b, and c and two things are drawn without replacement. The possible samples produced are ab, ac, and bc. There are three possible combinations. Let's do a problem where you should not try to enumerate the possible combinations. What if you want to determine how many committees of 4 can be formed from 7 people. There are PERMUT(7, 4) = 840 permutations possible sequences of 4 people formed from 7. For each combination of 4 has 4 times 3 times 2 times 1 = 4! permutations. So, there are 840 permutations divide (4! permutations/combination) = 35 committees of size 4 that can be formed from 7. Better screen some of the candidates first to reduce the sample space. In general, the number of combinations = PERMUT(N, n) divide n!, or doing it all with one Excel function, it's = COMBIN(N, n). Example: How many 5-card hands can be drawn from a 52 card deck? There are 2, 598, 960 possible hands. Can you compute it? How many committees of 5 people can be formed from 8 people?

Explanation / Answer

If we have to form committees of 5 people from 8 people

So the following subcommittees are all the same even though the ordering is different :

{1,2,3,4,5} , {5,4,3,2,1} , {2,3,5,1,4}

So now you can use the "n choose r" function :

  nCr = n! ÷ [ r! * (n - r)! ]

Now substitute 8 for n and 5 for r in the above equation and we get -

  nCr = 8! ÷ [ 5! * (8 - 5)! ]

= (8*7*6* 5!) ÷ (5! * 3!)

so 5! on numerator and denomenator gets cancelled out.

= (8 * 7 * 6)÷ (3 * 2 * 1)

= 56

Answer 56 committees can be formed.

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