Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.


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

share|improve this question
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 3 '13 at 4:54
@BrianM.Scott Does having it "less than" vs. "less than or equal to make a difference"? –  AlanH Feb 3 '13 at 5:27
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 3 '13 at 5:31

2 Answers 2

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."

share|improve this answer
Without the condition, it then becomes a weak composition, correct? –  AlanH Feb 3 '13 at 5:09
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 3 '13 at 5:13

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 $\binom{100-1}{5-1}$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.