Practice Problem 1

First, however we distribute the blanks, there will be $5! = 120$ ways to order the $5$ distinct letters.

Next, let $a_{n}$ be the number of blanks between the $n$ th and $(n+1)$ st letters for $1 \le n \le 4.$ We then require that $a_{1} + a_{2} + a_{3} + a_{4} = 15$ such that $a_{n} \ge 3$ for each $n.$

The number of solutions to this equation will be the same as the number of ways of arranging the blanks as required. Now let $a_{n}' = a_{n} - 3$ for each $n.$ Our equation the becomes

$a_{1}' + a_{2}' + a_{3}' + a_{4}' = 15 - 4*3 = 3$ , such that $a_{n}' \ge 0$ for each $n.$

This is now a stars and bars calculation, and thus has $\dbinom{3 + 4 - 1}{3} = \dbinom{6}{3} = 20$ solutions.

Thus the number of ways of arranging the letters and blanks as specified is $120*20 = \boxed{2400}.$

For $1\leqslant i \leqslant 4$ , let $x_{i}(\geqslant 3)$ , be the number of blanks between the ith and the (i+1)th lettter. Then

$x_{1} + x_{2} + x_{3} + x_{4} = 15 \cdots Eq. 1$

The Number Of Solutions of Eq. 1-

$=$ coefficient of $t^{15}$ in $(t^{3} + t^{4} + \cdots)$

$=$ coefficient of $t^{3}$ in $( 1 - t ) ^ {-4}$

$=$ coefficient of $t^{3}$ in $(1 + _{1}^{4}\textrm{C} + _{2}^{5}\textrm{C}t^{2} + _{3}^{6}\textrm{C}t^{3} \cdots)$

$= _{3}^{6}\textrm{C}$

$= 20$

But the 5 letters can be permuted in $5! = 120$ ways.

Thus the required number of arrangements $= 120.20$

$\boxed{2400}$