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
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
Updated On: Dec 6, 2024
- $2^7$
- $8.2^8$
- $10.2^7$
- None of these
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The Correct Option is C
Solution and Explanation
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 $C_{0 }+ 2C_{1} + 3C_{2} + ..... + 9C_{8}$
$= 2^{8} + 8.2^{7}. = 2^{7} \left(2 + 8\right) = 10.2^{7}.$
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Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.