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.
• B.
• C.
• D.

Answer 1

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

$({}^8{C}_0+{}^8{C}_{1}x+{}^8{C}_2{x}^2+..........+{}^8{C}_8{x}^8)\times (1+2x+3x^2+4x^3+.........+9x^8+.........)$

= $9.{}^8{C}_0+8.{}^8{C}_1+7.{}^8{C}_2+.............+1.{}^8{C}_8$

= $C_0+2C_{1}x+{}^3{C}_2{x}^2+......{}^9{C}_8{x}^8$ = $(1+x{)}^8+8x(1+x{)}^7$

Putting x = 1, we get $C_0+2C_1+3C_2+........+9C_8$

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