1

I'm currently studying a book by Paul Zeitz and currently stuck on exercise 6.2.23, below is the problem:

Find a formula for the number of different ordered triples $(a,b,c)$ of positive integers that satisfy $$a+b+c=n$$

For my first attempt I first started by looking at some values for $n$ like $n=50$ which by using 'Stars and Bars Technique' gets you to the solution $\binom{52}{2}$. I am wondering if the same technique could be applied to the question above so that I can obtain a formula for it.

Apologies if I am missing something elementary here.

RobPratt
  • 50,938
JAB
  • 639
  • Yes, Stars and Bars works for all $n$. This is literally asking for the Stars and Bars solution given that you have three positive terms which sum to $n$. – lulu Jun 30 '24 at 08:15
  • Does it mean that the solution is $\binom{n+2}{2}$? – JAB Jun 30 '24 at 08:17
  • 1
    Your problem requires positive solutions, not non-negative solutions. For positive solutions, you should be getting $\binom {n-1}2$. For $n=3$, say, the only solution is $(1,1,1)$. Your formula would give $\binom 52=10$ corresponding to the solutions $(1,1,1)$, the six permutations of $(2,1,0)$ and the $3$ permutations of $(3,0,0)$. if $0$ were allowed as a solution. – lulu Jun 30 '24 at 08:21
  • Ah I see, I should be using the $\binom{n-1}{k-1}$ formula right? Thanks for clearing things up! – JAB Jun 30 '24 at 08:25
  • 2
    As a general suggestion: always work explicit examples numerically to confirm formulas. With this sort of question, it's usually easy to write out small examples completely. Quick way to test formulas. – lulu Jun 30 '24 at 08:31
  • 2
    Do you see that there's a one-one correspondence between solutions of $a+b+c=n$ in positive integers and solutions of $a+b+c=n-3$ in nonnegative integers? – Gerry Myerson Jun 30 '24 at 09:03
  • Here is a similar question. – rtybase Jun 30 '24 at 19:08

0 Answers0