. asked in Science & MathematicsMathematics · 8 years ago

Give a combinatorial argument to show that...?

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

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  • 8 years ago
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    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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