2

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.

V.G
  • 4,196
Ahmed Maged
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  • If you know that ${10 \choose 2} = 45$, you don't need to use trial and error. – Toby Mak Feb 10 '21 at 09:20
  • If you consider that $ \ 45 \ = \ 5 \ · \ 3 \ · \ 3 \ $ , and are used to calculating binomial coefficients, you could ask how you might get such a product by canceling factors in products of consecutive integers. One way would be to have $ \ \frac{10 \ · \ 9}{2} \ $ , which would be the result of $ \ \frac{10!}{8! \ · \ 2!} \ \ . $ –  Feb 10 '21 at 09:23
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    @TobyMak But what about $\binom {45}1$? But the question which binomial coefficients are equal to $45$ reduces the search space significantly. – Mark Bennet Feb 10 '21 at 09:23

2 Answers2

1

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.

Joshua P. Swanson
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0

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 $$\binom {4+3r}{r}\le \binom {4+3r}{1}= 4+3r \le 7.$$

DanielWainfleet
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