Mathematics - Permutation & Combination Question with Solution | TestHub

Given, $C_1+C_2+C_3+C_4=30$

Now it has given that $4\le C_2\le 7 \& 2\le C_3\le 6$

and $C_1 \& C_4$ can take any value.

Finding coefficient of $x^{30}$ in

$\left(x^0+x^1+......x^{30}\right)\left(x^4+x^5+......x^7\right)\left(x^2+...x^6\right)\left(x^0+....x^{30}\right)$

$={\left(\frac{1-x^{31}}{1-x}\right)}^2{x}^4\left(\frac{1-x^4}{1-x}\right)x^2\left(\frac{1-x^5}{1-x}\right)$

{Ignoring higher power more than $30$}

$={\left(\frac{1}{1-x}\right)}^2{x}^4\left(\frac{1-x^4}{1-x}\right)x^2\left(\frac{1-x^5}{1-x}\right)$

$=x^6\left(1-x^4\right)\left(1-x^5\right){\left(1-x\right)}^{-4}$ $=x^6\left(1-x^5-x^4+x^9\right){\left(1-x\right)}^{-4}$

$=\left(x^6-x^{11}-x^{10}+x^{15}\right){\left(1-x\right)}^{-4}$

Required coefficient $={}^{4+24-1}C_{24}-{}^{4+19-1}C_{19}$ $- {}^{20+4-1}C_{20}+{}^{15+4-1}C_{15}$

$={}^{27}C_{24}-{}^{22}C_{19}-{}^{23}C_{20}+{}^{18}C_{15}$ $=430$