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I attempted this problem in many ways, but all of them were wrong. I started off using the stars and bars method, which I used to solve my previous problem.

  1. I had a total of fourteen stars as the solution contains integers less than 14. I put in three bars as I could divide the stars into 4 groups. Then I added the stars and bars (17) and did 17 choose 3 getting 680. This was my first attempt

  2. I attempted to count the solutions but I realized this was too tedious and moved on

  3. I realized my stars and bars were only for the number 14 and not 13, 12, 11, 10 and so on so I added the stars and bars And I got 3050 which is completely wrong.

I just don't know where to go and what theorem to use, an answer would be nice but any guidance would be really helpful.

Jyrki Lahtonen
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XXIMPXX
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  • Perhaps it is easiest to add a new non-negative variable $w=14-x-y-z-t$ and look at $x+y+z+t+w=14$. Then a routine application of Stars and Bars gives $\binom {14+5-1}{5-1}=3060$. – lulu Sep 24 '22 at 23:28
  • Note: your method $#3$ should have worked...$\sum_{n=0}^{14}\binom {n+4-1}{4-1}=3060$ as before. – lulu Sep 24 '22 at 23:34

1 Answers1

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To summarize the discussion on the comments:

Your third method certainly should have worked, but you appear to have made an arithmetic error somewhere. We have $$\sum_{n=0}^{14}\binom {n+4-1}{4-1}=3060$$

not $3050$ as you claim.

A quicker method: define a new non-negative variable $$w=14-x-y-z-t$$ and now consider the equality $$x+y+z+t+w=14$$

A routine application of Stars and Bars yields $$\binom {14-5-1}{5-1}=3060$$ as before.

lulu
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