## Tuesday, 22 November 2011

### Algebra-25 (IIFT-2009)

The number of distinct terms in the expansion of (X+Y+Z+W)30 is:
(1)4060                      (2)5456                      (3)27405                    (4)46376
Solution follows here:

Solution:
This is application of binomial theorem:
(x+y)n = nC0 xny0 + nc1 xn-1y +nC2 xn-2y2 +….+ nCn x0yn
Considering (X+Y) as first term and (Z+W) as second term observe the following binomial expansions:

(X+Y+Z+W)1                = 1C0 (X+Y)(Z+W)0 + 1C1 (X+Y)0(Z+W)
Number of terms:              2*1                         +      1*2

(X+Y+Z+W)2       = 2C0 (X+Y)2(Z+W)0 + 2C1  (X+Y)1(Z+W)1 + 2C2 (X+Y)0 (Z+W)2
No.of terms:                   3*1                +           2*2                      +       1*3

(X+Y+Z+W)3 = 3C0 (X+Y)3(Z+W)+ 3C1 (X+Y)2(Z+W) + 3C2 (X+Y)1(Z+W)2 + 3C3 (X+Y)0(Z+W)3
No.of terms:     4*1              +         3*2          +       2*3                +           1*4

If we observe this pattern, summary is:
(X+Y+Z+W)1 yields 2*1 + 1*2 terms
(X+Y+Z+W)2 yields 3*1 + 2*2 + 1*3 terms
(X+Y+Z+W)3 yields 4*1 + 3*2 + 2*3 +1*4 terms
Going by this logic,
(X+Y+Z+W)30 yields
31*1 + 30*2 + .....+ 17*15 + 16*16 +..........+ 1*31     terms
= 2{1*31 + 2*30 +....15*17} + 16*16
= 2{117 n(32-n)} + 16*16
= 2{(117 32*n) - (117 n2)} + 256
= 2{32(117 n) - (117 n2)} + 256 = 2{32(n(n+1)/2)n=17 - (n(n+1)(2n+1)/6)n=17} +256
= 5456   terms
Distinct number of terms in (X+Y+Z+W)30 = 5456