Factorial equations

Assume w.l.o.g. that $a \le b$ . Combining the two conditions gives: $\begin{aligned} a! &= {(a + (b-3))! \over b!} \\ &= (b+1)(b+2)....(b+(a-3)) \end{aligned}$

We see that $(a-3)$ must be at least $1$ (otherwise this product has no terms). The first few potential solutions are, for clarity: $\begin{aligned} 4! &= (b+1) \\ 5! &= (b+1)(b+2) \\ 6! &= (b+1)(b+2)(b+3) \\ 7! &= (b+1)(b+2)(b+3)(b+4) \\ 8! &= (b+1)(b+2)(b+3)(b+4)(b+5) \\ \end{aligned}$

Working through this list, we find that the $4!$ and $6!$ lines have solutions: $a=4, b=23, c=24 \\ a=6, b=7, c=10$ .

We can check that $5! = (b+1)(b+2)$ has no solution.

Now, when we get onto the $7!$ case and higher, we run into a problem. The minimum possible value for the r.h.s., which occurs when $b=a$ , is $(2a-3)! \over a!$ But for $a=7$ , this expression is already greater than $7!$ .

We can easily show by induction (with $a=7$ as the starting step) that $a! < {(2a-3)! \over a!}$ for all $a \ge 7$ : $\begin{aligned} (a!)^2 &\lt (2a-3)! \\ (a!)^2(a+1)^2 &\lt (2a-3)!(a+1)^2\\ &\lt (2a-3)!(2a-2)(2a-1) \\ &= (2a-1)!\\ (a+1)!^2 &\lt (2(a+1)-3)! \end{aligned}$

So we only have the two solutions we found earlier, $c=24$ plus $c=10$ gives $\boxed{34}$ .

From the second equation, we can get that $(a+b)! = (c+3)!$ , and dividing this by the first equation gives us $\binom{a+b}{a} = 6\binom{c+3}{3}$ . Let $a+b=c+3=n$ . The problem reduces to finding which rows of Pascal's triangle have $x$ and $6x$ in them for some positive integer $x$ .

Let $a=4$ . Then $6\binom{n}{3} = \binom{n}{4} \to \frac{(n-3)!}{(n-4)!} = 4! \to n-3 = 24 \to$ $c=24$ . Thus, $(a,b,c) = (4,23,24)$ satisfies the equations. Continuing in this manner (increasing $a$ by $1$ and seeing if there is a positive integer solution for $c$ ) gives us the following results:

$a = 5 \to$ $(c)(c-1) = 5! = 120 \to c \not\in \mathbb{Z}^+$

$a = 6 \to$ $(c)(c-1)(c-2) = 6! = 720 \to c = 10$

This last solution corresponds to the row $n = 13$ . We can quickly verify by inspection that there are no solutions for rows $n < 13$ , and there are clearly no other solutions for $n>13$ , since we essentially checked them above.

Thus, there are only two possible values for $c$ , namely $10$ and $24$ , and so our answer is $10 + 24 = \fbox{34}$ .

Without loss of generality, we can assume $a \geq b$ . To get a feel for the situation we try $a=c-1$ .

This immedeately gives $b=4$ from the second equation and $c=b!$ from the first. So $c=24$ is one of the possibilities we seek.

Next we try $a=c-2$ to get $b=5$ from the second equation and $c*(c-1)=b!$ from the first. Thus we get $120=c*(c-1)$ . Unfortunately this equation has no integer solutions.

Trying $a=c-3$ similarly gives the equation $720=c*(c-1)*(c-2)$ with solution $c=10$ (and $a=7, b=6$ ).

Trying $a=c-4$ similarly gives us the equation $5040=c*(c-1)*(c-2)*(c-3)$ with solution $c=10$ (and $a=6, b=7$ ).

Since we could assume $a \geq b$ , we have found all possibilities; $c=24$ and $c=10$ . This results in the answer $34$ .