# Give a combinatorial argument to show that...?

C(n,k) = C(n,n-k)

### 1 Answer

Relevance

- MathMan TGLv 78 years agoFavorite Answer
n items: 1 2 3 4 ..... k || (k+1) (k+2) . . . . n

........... ← chosen → || ← .. not chosen →

For every set of k items left of the bars,

there is a unique set of (n-k) items on the other side.

Thus, they are 1-to-1, and there is the same number of each.

For example, 5c2 = 5c3

1 2 || 3 4 5

1 3 || 2 4 5

1 4 || 2 3 5

1 5 || 2 3 4

2 3 || 1 4 5

2 4 || 1 3 5

2 5 || 1 3 4

3 4 || 1 2 5

3 5 || 1 2 4

4 5 || 1 2 3

10 of each.

Also in the formula:

since k = n - (n-k), we have:

C(n, k) = n! / (k! (n-k)! )

C(n, n-k) = n! / ((n-k)! (n - (n-k))! = n! / ( (n-k)! k! )

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