24!=a!*b!*c!*d! Find total number of ordered quadrapules which satisfy the given equation?

Question

I have found 92 as answer after trying possible quadrapules. Can anyone suggest different approach?

Answer 1

Let $m \ge n \ge p \ge q \ge 0$ be a reordering of $a, b, c, d$ in descending order. In order to reproduce the number $92$, $a,b,c,d$ is allowed to take the value $0$.

Since $23|24!$ and $23\!\!\mathop{\not|}k!$ for any $k < 23$, we have $m \ge 23$.

• If $m = 24$, there are $4$ possible ways to assign $m$ (i.e. $m = a$, $m = b$, $m= c$ or $m = d$ ). For each assignment of $m$, the remaining $3$ variables can take values of $0$ or $1$ independently. There are $4\times 2^3 = 32$ solutions with $m = 24$.

• If $m = 23$, then $n!p!q! = \frac{24!}{23!} = 24 = 4!$. Since $3|4!$ and $3\!\!\mathop{\not|}k!$ for any $k < 3$, we have $n \ge 3$.

  • If $n = 4$, there are $4$ possibles ways to assign $m$, $3$ possible ways to assign $n$ to remaining slot. For each assignment of $m,n$, the remaining $2$ variables take values of $0$ or $1$ independently. There are $4 \times 3 \times 2^2 = 48$ solutions with $m = 23, n = 4$.

  • Otherwise $n = 3$, then $p!q! = 4$ and it forces $p = q = 2$. In this case, there are $4$ possible ways to assign $m$ and $3$ possible ways to assign $n$. There are $4 \times 3 = 12$ solutions with $m = 24, n = 3$.

This exhausts all possibilities and there are totally $32 + 48 + 12 = 92$ solutions.

Answer 2

Guide to get you started:

One possible approach.

Suppose $a \le b \le c \le d$, we have $d \le 24$.

Case $1$: if $d=24$, then what can we say about $a,b,c$?

Case $2$: if $d \ne 24$, since $23|24!$ and $23$ is a prime, we need $d=23$ and we reduce the problem to $a!b!c!=24$. Under this scenario, we have $c \le 4$, repeat similar argument.

After you deduce possible values of $(a,b,c,d)$ that satisfies $a \le b \le c \le d$., note that the original question doesn't impose $a \le b \le c \le d$.