Question

Let "$C$" be the number of Combinations and "$P$" be the number of Permutations of $4$ letters taken from the word $EXAMINATION$. If the sum of digits of $C+P$ is $k$, then find the value of $\frac{k}{2}$.

Answer 1

There are $11$ letter $A,A,I,I,N,N,E,X,M,T,O$
Then the no. of combinations = coefficient of $x^4$ in $(1+x+x^2{)}^3(1+x{)}^5(\because 2A^{\prime }s,2I^{\prime }s,2N^{\prime }s,1E,1X,1M,1Tand1O)$
$=\text{Coefficient of}{x}^{4}in\left\{\right(1+x{)}^3+x^6+3(1+x{)}^2{x}^2+3(1+x\left)x^4\right\}(1+x{)}^5$
$=\text{Coefficient of}{x}^{4}in\left\{\right(1+x{)}^8+x^6(1+x{)}^5+3x^2(1+x{)}^7+3x^4(1+x{)}^6\}$
$={}^8{C}_4+0+3.{}^7{C}_2+3$
$=\frac{\mathrm{8.7.6.5}}{\mathrm{1.2.3.4}}+3.\frac{7.6}{1.2}+3$
$=70+63+3$
$=136$
And No. of permutations
$=\text{coefficient of}{x}^{4}in4!{(1+\frac{x}{1!}+\frac{x^2}{2!})}^3{(1+\frac{x}{1!})}^5$
$=\text{coefficient of}{x}^{4}in4!{(1+x+\frac{x^2}{2})}^3{(1+x)}^5$
$=\text{coefficient of}{x}^{4}in4!\left\{\right(1+x{)}^3+\frac{x^6}{8}+\frac{3}{2}(1+x{)}^2{x}^2+\frac{3}{4}{x}^4(1+x)\}(1+x{)}^5$
$=\text{coefficient of}{x}^{4}in4!\left\{\right(1+x{)}^8+\frac{x^6}{8}(1+x{)}^5+\frac{3}{2}{x}^2(1+x{)}^7+\frac{3}{4}{x}^4(1+x{)}^6\}$
$=4!\{{}^8{C}_4+0+\frac{3}{2}.{}^7{C}_2+\frac{3}{4}\}$
$=4!\left\{\frac{\mathrm{8.7.6.5}}{\mathrm{1.2.3.4}}+\frac{3}{2}.\frac{7.6}{1.2}+\frac{3}{4}\right\}$
$=\mathrm{8.7.6.5}+6\left(\mathrm{3.7.6}\right)+6.3$
$=1680+756+18$
$=2454$
$\therefore$ Sum of digit is $k=16$

Hence, $\frac{k}{2}=8$