Question

Closed form for $\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}$

score 128 · Answer 1 · 2014-04-09 17:49:29

128

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Closed form for $\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}$

Published on 09 Apr 2014 - 17:49

score 1k · Answer 2 · 2026-03-30 03:03:28

1k

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Proving combinatorial identity $\sum_{k=0}^{n}(-1)^k \binom{2n-k}k 2^{2n-2k} = 2n+1$ "directly"

Published on 30 Mar 2026 - 3:03

score 100 · Answer 3 · 2014-04-10 08:16:38

100

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How to prove this indentity $\binom{100}{0}^2-\binom{100}{1}^2+\binom{100}{2}^2-...-\binom{100}{99}^2+\binom{100}{100}^2=\binom{100}{50}$

Published on 10 Apr 2014 - 8:16

score 751 · Answer 4 · 2026-03-28 08:17:06

751

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Preventing "proof by homework"?

Published on 28 Mar 2026 - 8:17

score 377 · Answer 5 · 2014-04-12 17:36:40

377

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How to closed the sum $\displaystyle \sum_{k=0}^n \dfrac{(-1)^k(2k+1)!!}{(n-k)!k!(k+1)!}$

Published on 12 Apr 2014 - 17:36

score 84 · Answer 6 · 2014-04-13 02:50:21

84

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Sum of certain binomial coefficients

Published on 13 Apr 2014 - 2:50

score 231 · Answer 7 · 2014-04-13 17:04:04

231

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Cumulative Binomial Distribution function , Solve for n (trials)

Published on 13 Apr 2014 - 17:04

score 132 · Answer 8 · 2014-04-13 19:52:35

132

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Very loose bound on sum of first binomials

Published on 13 Apr 2014 - 19:52

score 22.8k · Answer 9 · 2026-04-03 01:01:54

22.8k

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Prove that $\sum_{k=0}^n k^2{n \choose k} = {(n+n^2)2^{n-2}}$

Published on 03 Apr 2026 - 1:01

score 88 · Answer 10 · 2026-03-28 08:09:44

88

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Show that if $a,k\in \mathbb{Z}$ with $0\leq k \leq a$, then $\binom ak=\frac{a!}{k!(a-k)!}=\binom {a}{a-k}$.

Published on 28 Mar 2026 - 8:09

score 45 · Answer 11 · 2014-04-14 06:55:04

45

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Finding the value of an expression involving co-efficients in binomial expansion

Published on 14 Apr 2014 - 6:55

score 108 · Answer 12 · 2014-04-14 10:33:35

108

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Set of solutions for a binomial inequality

Published on 14 Apr 2014 - 10:33

score 69 · Answer 13 · 2014-04-14 19:26:48

69

Views

Inequality with two binomial coefficients

Published on 14 Apr 2014 - 19:26

score 71 · Answer 14 · 2014-04-15 14:10:58

71

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Sum of Binomial Series of form $\binom{2000}{3k-1}$

Published on 15 Apr 2014 - 14:10

score 210 · Answer 15 · 2026-03-30 06:46:15

210

Views

Proof on Divisibility of Binomial Coefficients

Published on 30 Mar 2026 - 6:46

List Question

Closed form for $\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}$

Proving combinatorial identity $\sum_{k=0}^{n}(-1)^k \binom{2n-k}k 2^{2n-2k} = 2n+1$ "directly"

How to prove this indentity $\binom{100}{0}^2-\binom{100}{1}^2+\binom{100}{2}^2-...-\binom{100}{99}^2+\binom{100}{100}^2=\binom{100}{50}$

Preventing "proof by homework"?

How to closed the sum $\displaystyle \sum_{k=0}^n \dfrac{(-1)^k(2k+1)!!}{(n-k)!k!(k+1)!}$

Sum of certain binomial coefficients

Cumulative Binomial Distribution function , Solve for n (trials)

Very loose bound on sum of first binomials

Prove that $\sum_{k=0}^n k^2{n \choose k} = {(n+n^2)2^{n-2}}$

Show that if $a,k\in \mathbb{Z}$ with $0\leq k \leq a$, then $\binom ak=\frac{a!}{k!(a-k)!}=\binom {a}{a-k}$.

Finding the value of an expression involving co-efficients in binomial expansion

Set of solutions for a binomial inequality

Inequality with two binomial coefficients

Sum of Binomial Series of form $\binom{2000}{3k-1}$

Proof on Divisibility of Binomial Coefficients

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