Question

Inclusion-exclusion conundrum: $\sum_{k=1}^{\lfloor n/2\rfloor} (-1)^{k-1}k\binom{n-k}{k}2^{n-2k} = \binom{n+1}{3}$

score 107 · Answer 1 · 2026-04-16 13:11:16

107

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Inclusion-exclusion conundrum: $\sum_{k=1}^{\lfloor n/2\rfloor} (-1)^{k-1}k\binom{n-k}{k}2^{n-2k} = \binom{n+1}{3}$

Published on 16 Apr 2026 - 13:11

score 399 · Answer 2 · 2026-04-11 08:25:41

399

Views

Does this combinatorial identity hold？

Published on 11 Apr 2026 - 8:25

score 81 · Answer 3 · 2026-04-17 01:45:02

81

Views

Variant of identity for a sum of binomial coefficient (Hockey-stick)

Published on 17 Apr 2026 - 1:45

score 123 · Answer 4 · 2026-04-12 22:13:22

123

Views

summation of a weighted product of binomial coefficients

Published on 12 Apr 2026 - 22:13

score 42 · Answer 5 · 2026-04-17 13:46:15

42

Views

Prove that the 5-adic valuation of $(10^{k+t}+10^t+1)^c-1$ is $t+v_5(c)$

Published on 17 Apr 2026 - 13:46

score 405 · Answer 6 · 2026-04-11 01:29:52

405

Views

Bivariate generating function for squared binomial coefficients

Published on 11 Apr 2026 - 1:29

score 117 · Answer 7 · 2026-04-12 10:16:42

117

Views

Derivation for identities involving Stirling Numbers and Binomial Coefficients

Published on 12 Apr 2026 - 10:16

score 38 · Answer 8 · 2026-04-12 01:22:19

38

Views

How do we solve this cumulative question involving sequences and binomial coefficients

Published on 12 Apr 2026 - 1:22

score 301 · Answer 9 · 2026-04-15 13:00:54

301

Views

I need the combinatorial proof for the following identity

Published on 15 Apr 2026 - 13:00

score 159 · Answer 10 · 2026-04-17 14:17:29

159

Views

Is $\sum_{s=n}^{m}\binom{m}{s}(x-1)^{m-s}=\sum_{s=n}^m\sum_{j=s}^m\binom{s-1}{n-1}\binom{m-s}{j-s}(x-1)^{m-j+s-n}$ an identity?

Published on 17 Apr 2026 - 14:17

score 115 · Answer 11 · 2026-04-16 12:51:27

115

Views

How to prove $\sum_{k=p}^n (-1)^k\binom{n}{k}\binom{k}{p} = (-1)^p$ by using generating functions

Published on 16 Apr 2026 - 12:51

score 155 · Answer 12 · 2026-04-13 10:46:28

155

Views

Combinations with repetition? $n+r-1 \choose r-1$=$n+r-1 \choose r$

Published on 13 Apr 2026 - 10:46

score 47 · Answer 13 · 2026-04-11 13:16:47

47

Views

The number of binary sequences of length 300 made up of 50 ones and 250 zeros so that between Every two ones has at least one zero.

Published on 11 Apr 2026 - 13:16

score 70 · Answer 14 · 2026-03-26 04:36:21

70

Views

Number of combinations of size $k$ of elements in a multiset with finite multiplicites

Published on 26 Mar 2026 - 4:36

score 69 · Answer 15 · 2026-04-15 18:11:37

Found a pattern on Pascal's triangle: ${n \choose r}=\sum_{i=0}^k {k \choose i}{n-k \choose r-k+i}$. Has anyone come across it anywhere?

List Question

Inclusion-exclusion conundrum: $\sum_{k=1}^{\lfloor n/2\rfloor} (-1)^{k-1}k\binom{n-k}{k}2^{n-2k} = \binom{n+1}{3}$

Does this combinatorial identity hold？

Variant of identity for a sum of binomial coefficient (Hockey-stick)

summation of a weighted product of binomial coefficients

Prove that the 5-adic valuation of $(10^{k+t}+10^t+1)^c-1$ is $t+v_5(c)$

Bivariate generating function for squared binomial coefficients

Derivation for identities involving Stirling Numbers and Binomial Coefficients

How do we solve this cumulative question involving sequences and binomial coefficients

I need the combinatorial proof for the following identity

Is $\sum_{s=n}^{m}\binom{m}{s}(x-1)^{m-s}=\sum_{s=n}^m\sum_{j=s}^m\binom{s-1}{n-1}\binom{m-s}{j-s}(x-1)^{m-j+s-n}$ an identity?

How to prove $\sum_{k=p}^n (-1)^k\binom{n}{k}\binom{k}{p} = (-1)^p$ by using generating functions

Combinations with repetition? $n+r-1 \choose r-1$=$n+r-1 \choose r$

The number of binary sequences of length 300 made up of 50 ones and 250 zeros so that between Every two ones has at least one zero.

Number of combinations of size $k$ of elements in a multiset with finite multiplicites

Found a pattern on Pascal's triangle: ${n \choose r}=\sum_{i=0}^k {k \choose i}{n-k \choose r-k+i}$. Has anyone come across it anywhere?

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