Let "A" be an n-element set and let k E N. How many functions f : A --> {0,1} are the for which there are exactly k elements in "A" with f(a)=1 ? Note: E = "be a member of" (k E N) N = Natural numbers
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If $\displaystyle k>n$ the answer is 0. If $\displaystyle 1\le k \le n$ then the answer is $\displaystyle \binom{n}{k}$. You may think of the number of ways to arrange k 1's and (n-k) 0's.
Thank you Plato!
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