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

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) 27 (B) 8.28 (C) 10.27 (D) None of these

Tardigrade Admin · Accepted Answer

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{( 1 - x ^{9} ) ^{2}}{( 1 - x ) ^{2}} (1 + x)^{8}

= coefficient of

x^{8} in (1 + x)^{8} in (1 + x^{8}) (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 \cdot^{8} C_{1} + 7 .^{8} C_{2} + .... + 1 .^{8} C_{8}

= C_{0} + 2 C_{1} + 3 C_{2} + .... + 9 C_{8} [C_{r} =^{8} C_{r}]

Now

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

Differentiating with respect to x, we get

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} (2 + 8) = 10. 2^{7} .

Q. 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

$2^{7}$

$8. 2^{8}$

$10. 2^{7}$

None of these