Question

$15$ identical jewels are to be distributed between $P,Q,R$ and $S$. Find the number of ways in which $P$ gets a maximum of $5$ and $Q$ gets at least $2$ jewels.

• A. $560$
• B. $120$
• C. $440$
• D. $680$

Answer 1

The correct option is C $440$

$P$ can't have more than $5$.
$Q$ should have at least $2$ and there are no restrictions on $R$ and $S$.
Coefficient of $x^{15}$ in
$(x^0+x^1+\cdots +x^5)(x^2+x^3+\cdots +x^{\infty })(x^0+x^1+\cdots +x^{\infty }{)}^2$
The above expression is forming G.P. series.
Therefore, we can simplify the above expression as $\left(\frac{1-x^6}{1-x}\right)\left(\frac{x^2}{1-x}\right){\left(\frac{1}{1-x}\right)}^2$

On taking $x^2$ common, we have to evaluate the coefficient of $x^{13}$ in $(1-x^6)(1-x{)}^{-4}$

$(1+x{)}^{-n}=\sum _{k=0}^{\infty }(-1{)}^k{}^{n+k-1}{C}_k{x}^k$
$\Rightarrow (1-x{)}^{-n}=\sum _{k=0}^{\infty }(-1{)}^{2k}{}^{n+k-1}{C}_k{x}^k$

Coefficient of $x^{13}$ in $(1-x{)}^{-4}={}^{4+13-1}{C}_{13}={}^{16}{C}_{13}={}^{16}{C}_3$
Coefficient of $x^7$ in $(1-x{)}^{-4}={}^{10}{C}_7={}^{10}{C}_3$

Therefore, coefficient of $x^{13}$ in $(1-x^6)(1-x{)}^{-4}={}^{16}{C}_3-{}^{10}{C}_3=440$