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How many positive integer solutions are there to $x_1 + x_2 + x_3 + x_4 < 100$?

I haven't seen any problems with "less than", so I'm a bit thrown off. I'm not sure if my answer is correct, but if there is, there has a be a more concise form.

Solution:

$$ \sum_{i=0}^{95} \binom{(99-i)+4-1}{99-i} $$

TheRealFakeNews
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    See my answer to this question; only a very small modification is needed. In brief, the trick is to count solutions to $x_1+x_2+x_3+x_4+x_5=100$, where $x_5$ takes up the slack between the solution to the inequality and the solution to the equation. – Brian M. Scott Feb 03 '13 at 04:54
  • @BrianM.Scott Does having it "less than" vs. "less than or equal to make a difference"? – TheRealFakeNews Feb 03 '13 at 05:27
  • A small one. So does the fact that in that problem we were counting non-negative solutions instead of positive solutions. The $<$ here actually makes it easier for you: positive solutions to the equation that André and I gave correspond to positive solutions of the original inequality. If you were counting non-negative solutions to the inequality, the righthand side of the equation would be $99$ instead. – Brian M. Scott Feb 03 '13 at 05:31

2 Answers2

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Hint: It is the number of positive solutions of $x_1+x_2+ x_3+x_4 +x_5=100$.

For in how many ways can I distribute candies among $4$ kids, each kid getting one candy at least, and with $\lt 100$ candies distributed?

Imagine I have $100$ candies. I call myself the fifth kid, and if $k$ candies are distributed among the real four, I get the remaining $100-k$. This gives a natural one to one correspondence between distribution of $\lt 100$ candies among $4$ kids, one at least to each, and distributions of $100$ candies among $5$ kids, at least one to each.

Mild modification of the idea takes care of the situation in which we do not have the condition "at least one to each."

André Nicolas
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  • Without the condition, it then becomes a weak composition, correct? – TheRealFakeNews Feb 03 '13 at 05:09
  • That's a term some people use. One can count them in basically the same "Stars and Bars" way. To distribute $n$ candies among $k$ kids, where some may get nothing, distribute $n+k$ candies, at least one to each, then take away a candy from everybody. – André Nicolas Feb 03 '13 at 05:13
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Using the hint provided by Andre Nicolas and Brian M. Scott, we add an extra variable $x_5$, then the problem can be viewed as asking for the number of compositions of $n=100$ into $k=5$ parts, or $\binom{100-1}{5-1}$.

TheRealFakeNews
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