Question: "C(n,r)=C(r,r)⋅C(n−r,0)+C(r,r−1)⋅C(n−r,1)+C(r,r−2)⋅C(n÷r,2)+C(r,r−3):C(n−r,3)++C(r,1)⋅C(a−r,r−1)+C(r,0)⋅C(n−r,r)"

Brennan Flores

Answered question

2020-12-25

$C (n, r) = C (r, r) \cdot C (n - r, 0) + C (r, r - 1) \cdot C (n - r, 1) + C (r, r - 2) \cdot C (n \div r, 2) + C (r, r - 3) : C (n - r, 3) + + C (r, 1) \cdot C (a - r, r - 1) + C (r, 0) \cdot C (n - r, r)$

Answer & Explanation

Arham Warner

Skilled2020-12-26Added 102 answers

Suppose that you have n objects, and you line them up. You need to pick r objects. If you take r objects from the first r objects, then you need to take 0 objects from the rest of the objects, which you have $n — r$ . Thus, you have $O (r, r) - C (n — r, 0)$ ways of doing this.
Similarly, if you pick k objects from the first r objects, you need to take $r — k$ objects from the rest of the objects. You have $O (r, k) - C (n — r, r — k)$ ways of doing this.
Since you can take Q, 1, ..., r objects from the first r objects, we have that $0 \leq k \leq r$ . Finally, since we can also pick k, this means that we can pick r objects from the n objects in $r \sum C (r, k) * C (n - r, r - k) k = 0$
ways. Thus we have proven that $r C (n, r) = \sum C (r, k) * C (n - r, r - k) k = 0$
which we needed to prove.

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