Page 166 - Fixtures

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∑ ∑ ∑. n n
n
(k + 1)C (n,k) = kC (n,k) = 2n + n2n−1
C (n,k) +

k =0 k =0 k =1

.( )

(x + y + z)12 x 3y5z 4 ()

(a + z)12 z4 a =x +y
a8 = (x + y)8 x 3y5
, .C(12, 4)a8z 4

x 3y5z 4 .C(8, 5)x 3y5

.C(12, 4) ×C (8, 5) = 12!

3!!× 4!!× 5!

()

2 × 1 ×C (n, 2) + 3 × 2 ×C(n, 3) + 4 × 3 ×C(n, 4) +

+ n(n − 1)C (n,n) = n (n − 1)2n−2

kC(n,k) = nC(n − 1,k − 1)
2 ×1×C(n, 2) = nC(n − 1,1) = n(n − 1)C(n − 2, 0)
3 × 2 ×C(n, 3) = 2nC(n − 1, 2) = n(n − 1)C(n − 2,1)
4 × 3 ×C(n, 4) = 3nC(n − 1, 3) = n(n − 1)C(n − 2, 2)

n(n − 1)C(n,n) = n(n − 1)C(n − 1,n − 1) = n(n − 1)C(n − 2,n − 2)