If $\binom{n}{r} = 45$, and $(n-3r)! = 24$, Is there a method of knowing $n$ and $r$ without guessing? I got the equation:
$n = 4 + 3r.$
So, $\displaystyle\binom{4+3r}{r} = 45$, and by using trial and error, I get $r = 2$, and $n = 10$, which is correct.
 A: Here's a very elementary approach. Roughly speaking, the entries in Pascal's triangle grow very quickly, except the $1$'s on the edges, so we can just compute enough of the triangle to be sure we've found all the $45$'s:
                      1
                    1   1
                  1   2   1
                1   3   3   1
              1   4   6   4   1
            1   5  10  10   5   1
          1   6  15  20  15   6   1
        1   7  21  35  35  21   7   1
      1   8  28   *   *   *  28   8   1
    1   9  36   *   *   *   *  36   9   1
  1  10  45   *   *   *   *   *  45  10   1
1  11   *   *   *   *   *   *   *   *  11   1

Here the *'s indicate the value is $ > 45$. The only 45's here are $\binom{10}{2} = \binom{10}{8} = 45$, and only the former is of the correct form. The only other 45's in the triangle will be $\binom{45}{1} = \binom{45}{44} = 45$, neither of which is of the correct form.
A: If $r\ge 2$ then $$\binom {4+3r}{r}= \prod_{j=0}^{r-1}\frac {(4+3r-j)}{j+1}\ge \frac {(4+3r)}{1}\cdot  \frac {(3+3r)}{2}\ge\frac {(10)(9)}{2}=45$$ (....because each term $\frac {(4+3r-j)}{j+1}$ is $\ge 1$ ....)
with equality only when $r=2$.
If $r\le 1$ then $$\binom {4+3r}{r}\le \binom {4+3r}{1}= 4+3r \le 7.$$
