Pascal's identity- counting

Question

+1

2313

2

Compute

$\binom50+\binom51+\binom62+\binom71+\binom83+\binom92+\binom{10}4+\binom{11}3$

.

Guest Apr 21, 2017

Guest · Answer 1 · 2017-04-21T11:55:11

nvm I got it. It's 23426. fourth number of the fifty third row of pascals triangle.

CPhill · Answer 2 · 2017-04-22T12:49:44

Using the identity : C(m, n) + C(m, n + 1) = C(m + 1, n + 1), we have

C(5,0) + C(5,1) + C(6,2) + C(7,1) + C(8,3) + C(9,2) + C(10,4) + C(11,3)

C(6,1) + C(6,2) + C(7,1) + C(8,3) + C(9,2) + C(10,4) + C(11,3)

C(7,2) + C(7,1) + C(8,3) + C(9,2) + C(10,4) + C(11,3)

C(7,1) + C(7,2) + C(8,3) + C(9,2) + C(10,4) + C(11,3)

C(8,2) + C(8,3) + C(9,2) + C(10,4) + C(11,3)

C(9,3) + C(9,2) + C(10,4) + C(11,3)

C(9,2) +C(9,3) + C(10,4) + C(11,3)

C(10,3) + C(10,4) + C(11,3)

C(11,4) + C(11,3)

C(11,3) + C(11,4)

C(12,4) =

495