Expert: Josh - 8/11/2006

Question
There are 8 players on a basketball team. How many different 5 man taems are possible if each player can play any position.  6th grade and stumped.  Can you help?

Answer
Hi Jack,

This particular question falls within a broader class of questions, where you have to work out the number of ways to select n items from N items. Let me introduce the basic ideas, so that you understand what this is about and hopefully can make some sense of the formula.

SECTION 1: The result

If we are only interested in finding the number of possible combinations, when we select n items from N items without replacement, and the order of appearance does not matter, we can "pick n from N" in C(N,n)=N!/[(N-n)!*n!] ways ....let's call this formula [#1]

where N!=N*(N-1)*(N-2)*...*1 is called the "N factorial".

So, for example, n!=n*(n-1)*(n-2)*(n-3)*...*2*1. When N=8, n=5, N!=8!=8*7*6*5*4*3*2*1=40320; (N-n)!=3!=3*2*1=6 and n!=5!=5*4*3*2*1=120.

Applying formula in [#1], you can form C(8,5)=8!/[(3)!*5!]=8*7*6*5*4*3*2*1/[3*2*1*(5*4*3*2*1)]=56 teams.

SECTION 2: Why the formula C(N,n)=N!/[(N-n)!*n!] works?

The rationale behind this is best explained with diagram.
We begin with eight people,
# # # # # # # #
Naturally, if we pick one person in random, this gives rise to eight different possibilities. If the person is selected without replacement (which makes sense in this problem), we are left with seven people after the first round of selection. We have
# # # # # # #
Now, if we pick the second person in random from the seven remaining candidates, there are seven possibilities. Again, the person is removed from the pool without replacement. We are left with six people after the second round of selection. We have
# # # # # #
Following this argument, since each member is independently selected, the total number of "permutations" (i.e., unique ways of empaneling the team) must be given by 8*7*6*5*4. ......[#2]

However, since we are only interested in the number of combinations that we can get, the order of appearance does not matter. This suggests that [#2] is over-counting. The main observation is that the same combination can appear in 5!=120 different ways, in various order.

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To understand this point. Let us consider a simple situation. Suppose person A, B, and C must be included in a team of three, the selection can proceed in any of the following order: A,B,C; A,C,B; B,A,C; B,C,A; C,A,B; or C,B,A.
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Therefore, the number of combinations is given by (8*7*6*5*4)/5!, which is equivalent to (8!/3!)/5! as suggested by formula [#1].

Some common values worth commiting to memory (because we use these a lot) are:
0!=1 (by definition)
1!=1
2!=2
3!=6
4!=24
5!=120
6!=720
7!=5040
8!=40320
9!=362880

Subsequent values may be found noting that n!=n*(n-1)!
For instance, 10!=10*9!=3628800.

The formula aside, I hope you get something out of the explanation offered in Section 2.

Let me know if you have any question.

Cheers:)