Compute \(\binom50+\binom51+\binom62+\binom71+\binom83+\binom92+\binom{10}4+\binom{11}3\) using Pascals Identity.
+128473 The identity is
C(n, k) = C(n - 1, k) + C(n - 1, k -1)
So 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(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(7,2)+ C(7,1)] + 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(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(9,3)+ C(9,2) ] + C(10,4) + C(11,3)
[C(10,3 ) ] + C(10,4) + C(11,3)
[C(10,3 ) + C(10,4)] + C(11,3)
[C(11,4) ] + C(11,3) =
C(12.4) =
495
