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# 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) # 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)

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Probability asked 2020-12-25
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)

## Answers (1) 2020-12-26
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
ways. Thus we have proven that PSKr C(n,r)=∑ C(r,k)*C(n-r,r-k) k=0ZSK
which we needed to prove.

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