Mathematics - Permutation & Combination Question with Solution | TestHub

For $1\le i\le 4$, let $x_i(\ge 3)$ be the number of blanks between the $i^{th}$ and $(i+1{)}^{th}$ letters. Then $x_1+x_2+x_3+x_4=15$. The number of solutions of (1) = coefficient of $x^{15}$ in ${\left(x^3+x^4+\dots \dots \right)}^4$.

$$\begin{array}{rl}& =\text{ coeff of }{x}^3\text{ in }(1-x{)}^{-4}\\ & =\text{ coeff of }{x}^3\text{ in }\left[1+4c_{1}x+5c_2{x}^2+6c_3{x}^3+\dots .\right]\\ & ={}^6{c}_3\\ & =20\end{array}$$

But 5 letters can be permuted in $5!=120$ ways. $\Rightarrow$ Required number of arrangements $=(120)(20)=2400$.