+0

# Pascal's identity- counting

0
161
2

Compute

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

.

Guest Apr 21, 2017
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#1
0

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

Guest Apr 21, 2017
#2
+77147
+1

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

CPhill  Apr 22, 2017

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