0

A more general formulation of the question would be: In how many ways can you permutate three natural numbers (not counting 0) such that their sum equals n?

Im also supposed to solve the problem in about 5 minutes with only pen and paper.

I know there are similar posts like: Given an integer $n >0$, how many ways can we express $n$ as the sum of three natural numbers $n_1,n_2,n_3$ ?

But I am having a hard time applying this to my problem since they are counting with zeros. The user "kneidell" comes with a good explanation of why the number of combinations counting with zeros is $\binom{n+2}{2}$.

One could just use the formula and subtract every combination who contains a zero. But I think thats a very ineffective way to do it. I tried using his logic to come up with my own formula or thinking process but failed.

So does anyone have an effective way to solve this question/these types of questions?

Stephen21212
  • 41
  • 4
    $\binom{50-1}{2}$... Effectively, for the 49 gaps between pages... pick where to place bookmarks indicating where you start/stop each session. – JMoravitz Oct 13 '22 at 13:49
  • Finding the number of tuples $(n_1,n_2,n_3)$ where the $n_i$ are positive integers that satisfy $n_1+n_2+n_3=50$ is practically the same as finding the number of tuples $(m_1,m_2,m_3)$ where the $m_i$ are nonnegative integers that satisfy $m_1+m_2+m_3=50-3=47$. If $(m_1,m_2,m_3)$ is a solution for the second problem then $(n_1,n_2,n_3)=(m_1+1,m_2+1,m_3+1)$ is a solution for the first problem. – drhab Oct 13 '22 at 13:55
  • The number of solutions to $$x_1 + x_2 + x_3 = n ~: x_1,x_2,x_3 \in \Bbb{Z_{\geq 1}}$$ is the same as the number of solutions to $$y_1 + y_2 + y_3 = (n-3) ~: ~y_1, y_2, y_3 \in \Bbb{Z_{\geq 0}}.$$ Simply set $y_i = x_i - 1 ~: ~i \in {1,2,3}.$ For Stars and Bars theory, see this article and this article. – user2661923 Oct 13 '22 at 13:57

0 Answers0