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
Options:
A) $ 2^{7} $
B) $ 8\ .\ 2^{8} $
C) $ 10\ .\ 2^{7} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- 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})}{{{(1-x)}^{2}}}}^{2}}{{(1+x)}^{8}} $ = coefficient of $ x^{8} $ in $ {{(1+x)}^{8}}{{(1-x)}^{-2}} $ = coefficient of $ x^{8} $ in $ {{(}^{8}}C_0{{+}^{8}}C_1x{{+}^{8}}C_2x^{2}+…….+{{}^{8}}C_8x^{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+3C_2+…..+9C_8 $ $ [C _{r}{{=}^{8}}C _{r}] $ Now $ C_0x+C_1x^{2}+…….+C_8x^{9}=x{{(1+x)}^{8}} $ Differentiating with respect to $ x $ , we get $ C_0+2C_1x+3C_2x^{2}+…9C_8x^{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}.(2+8)=10\ .\ 2^{7} $ .
BETA
