Question

Number of ways of selection of 8 letters from 24 letters of which 8 are a, 8 are b and the rest unlike, is given by

• A. $2^7$
• B. $8.2^8$
• C. $10.2^7$
• D. None of these

Answer 1

Answer: (C)

The number of selection = coefficient of $x^8$ in
$(1 + x + x^2 + .... + x^8) (1 + x + x^2 + ...... + x^8). (1 + x)^8$
= coefficient of $x^8$ in $\frac{\left(1-x^{9}\right)^{2}}{\left(1-x\right)^{2}} \left(1+x\right)^{8}$
= coefficient of $x^{8} in \left(1 + x\right)^{8} in \left(1 + x^{8}\right) \left(1 - x\right)^{-2}$
= coefficient of $x^{8}$ in
$\left(^{8}C_{0} + ^{8}C_{1}x + ^{8}C_{2}x^{2} + ..... + ^{8}C_{8}x^{8}\right)$
$× \left(1 + 2x + 3x^{2} + 4x^{3} + ..... + 9x^{8} +....\right)$
$= 9. ^{8}C_{0} + 8 · ^{8}C_{1} + 7. ^{8}C_{2} + .... + 1. ^{8}C_{8}$
$= C_{0} + 2C_{1 }+ 3C_{2 }+ .... + 9C_{8} \left[C_{r}= ^{8}C_{r}\right]$
Now $C_{0}x + C_{1}x^{2} + .... + C_{8}x^{9} = x \left(1 + x\right)^{8}$
Differentiating with respect to x, we get
$C_{0} + 2C_{1}x + 3C_{2}x^{2} + .... 9C_{8}x^{8} = \left(1 + x\right)^{8} + 8x \left(1 + x\right)^{7}$
Putting $x = 1$, we get $C_{0 }+ 2C_{1} + 3C_{2} + ..... + 9C_{8}$
$= 2^{8} + 8.2^{7}. = 2^{7} \left(2 + 8\right) = 10.2^{7}.$