The arrangement a (n, m) = n× (n- 1). (n-m+ 1) = n！ /(n-m)！ (n is subscript and m is superscript, the same below)

Combination C(n, m)=P(n, m)/P(m, m) =n! /m！ (noun) noun.

P is a permutation, the right foot code is n, and the right foot code is m, n (n-1) (n-2) ... (n-k+1);

C is combination, right foot code n, right foot code m, n (n-1) (n-2) ... (n-k+1)/m!

Extended data:

Let C(n- 1, k) and C(n- 1, k- 1) be odd numbers:

Have: (n-1)&; k = = k

(n- 1)； (k- 1)= = k- 1；

Since the last bit of k and k- 1 are necessarily different, the last bit of n- 1 must be 1.

Now suppose n &；; k == k .

Then the last bit of k is 1, because the last bit of n- 1 is different.

Because the last bit of n- 1 is 1, the last bit of n& is 0; k！ = k, which contradicts the hypothesis.

So get w-; k！ = k .

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