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

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$8.2^7$

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Solution

The correct option is C

The number of selections = 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 $\frac{(1 - x^{9})^{2}}{(1 - x)^{2}} (1 + x)^{8}$

= coefficient of $x^{8}$ in $(1 + x)^{3} (1 - x)^{- 2}$

= coefficient of $x^{8}$ in

$(^{8} C_{0} +^{8} C_{1} x +^{8} C_{2} x^{2} + . . . . . . . . . . +^{8} C_{8} x^{8}) \times (1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . . . . . + 9 x^{8} + . . . . . . . . .)$

= $9.^{8} C_{0} + 8.^{8} C_{1} + 7.^{8} C_{2} + . . . . . . . . . . . . . + 1.^{8} C_{8}$

= $C_{0} + 2 C_{1} x +^{3} C_{2} x^{2} + . . . . . .^{9} C_{8} x^{8}$ = $(1 + x)^{8} + 8 x (1 + x)^{7}$

Putting x = 1, we get $C_{0} + 2 C_{1} + 3 C_{2} + . . . . . . . . + 9 C_{8}$

= $2^{8} + {8.2}^{7}$ = $2^{7} . (1 + 8)$ = ${10.2}^{7}$ .

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