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Let's say we have $4$ positive integers $x$, $y$, $z$, $w$ that satisfy the property： $$x+ y+ z+ w = 100$$ How many possible distinct answers are there?

Xander Henderson
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A.Y
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    Welcome to stackexchange. To get help here you should get started and show us where you are stuck. Can you solve the problem if you try it with a number much smaller than $100? With fewer than four summands? – Ethan Bolker Nov 17 '19 at 02:54
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    It seems like it should be a very similar problem to count ways of picking 3 different numbers a<b<c< 100. Then you could let w=100-c, z= c-b, y = b-a, and x = a. – Robo300 Nov 17 '19 at 02:54
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    @Robo300 It's way easier to just stars and bars here. The answer is just $$99\choose3$$ – Rushabh Mehta Nov 17 '19 at 02:56
  • As you have not specified that the integers must be nonnegative, there are infinitely many solutions, no? – Xander Henderson Nov 17 '19 at 03:12
  • @XanderHenderson i originally answered that way. Then I reread the title. – David P Nov 17 '19 at 03:13
  • @DavidPeterson Oi... this is why vital parts of the question should be in the question, and not the title alone. – Xander Henderson Nov 17 '19 at 03:15

2 Answers2

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I'm going to write up a quick answer since this will be closed soon, but by stars and bars, we first assign $4$ of the $100$ to each variable to ensure that they will be positive. So, we have $96$ indistinguishable objects to distribute in $4$ buckets, giving us $${96+4-1\choose4-1}={99\choose3}=\color{red}{156849}$$

Rushabh Mehta
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Here is a direct approach:

For each choice of $x$, we count the number of sums $y+z+w=97-x$.

For each choice of $x$ and $y$, we count the number of sums $z+w=98-x-y$.

For each choice of $x$, $y$ and $z$, there is only one such $w=99-x-y-z$.

$$\sum_{x=1}^{97}\sum_{y=1}^{98-x}\sum_{z=1}^{99-x-y} 1 = \sum_{x=1}^{97}\sum_{y=1}^{98-x}(99-x-y) = 156849$$

David P
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