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So I have a question about a practice problem employing combination in which the repetition of elements is allowed, here it is:

Determine the number of non-negative integer solutions to the equation $a + b + c + d + e = 70$.

So this seems straightforward enough to me. I see 70 possible elements with 5 combinations, so $n=70$ and $r=5$. So employing the formula $C(n + r - 1, r) = C(n + r - 1, n - 1)$, I reach the following:

$C(70 + 5 - 1, 70 - 1) = C(74,69) = \frac{74!}{69!(74-69)!} =

\frac{74 \cdot 73 \cdot 72 \cdot 71\cdot 70}{5!} = 1.6108764 \cdot 10^{7}$

However, here is the answer supplied:

This is a combinations with repetition question without any special circumstances. There are 5 distinct objects and we are choosing exactly 70 of them. The number of ways to do this is C(70+5-1, 5-1) = C(74,4) = C(74, 70). (Grading: Recognizing combinations with repetition is worth 3 pts. Stating both 70 and 5 (or 4 depending on how they conceive of the problem) is worth 1 pt each. Applying the formula correctly is worth 5 pts

Could someone explain to me what I am doing wrong?

David Kohler
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BrotherJack
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  • You may want to look at http://math.stackexchange.com/questions/11468/if-there-are-200-students-in-the-library-how-many-ways-are-there-for-them-to-be/11475#11475. –  Jun 22 '11 at 03:00
  • One way to see it's 5 objects and 70 choices is: you have 70 balls; place them in 5 buckets, one labeled $a$, one labeled $b$, one labeled $c$, and labeled $d$, one labeled $e$. When you are done, $a$ is the number of balls in bucket $a$, $b$ is the number of balls in bucket $b$, etc. You have 5 possibilities for each ball, and you have to make the choices 70 times. How do you try to interpret "70 elements, five combinations"? – Arturo Magidin Jun 22 '11 at 03:15

4 Answers4

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Others have told you how to do it right, or have pointed to sites that tell you how to do it right. But I think the first thing is to be convinced you've done it wrong, and the easiest way to do that is to look at a much simpler example, say, $a+b=1$. Your reasoning gives 1 element with 2 combinations, so $n=1$ and $r=2$, so $C(2,2)=1$. But this is manifestly wrong, since there are 2 solutions, $a=1$, $b=0$, and $a=0$, $b=1$. Once you see why it's wrong, you're on your way to doing it right.

Gerry Myerson
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Please wee the Wikipedia article on Stars and Bars.

Questions very similar to this one have been asked on StackExchange a number of times, so a search will reveal quite a few detailed answers. But the Wikipedia article is longer than any of these answers, and very detailed.

André Nicolas
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  • Thanks User6312! I had seen a professor use this before, but he wasn't very clear on what he was doing at the time (or I was too slow to get it, or both!). This page was very helpful. – BrotherJack Jun 23 '11 at 20:07
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When you say it is obviously $C(70,5)$ what did you see?

I think a good way yo do a problem like this is to imagine that you have 70 balls on a line that you want to divide in 5 groups. To do that, you have to place 4 markers in between the balls to divide it in 5 groups. If you didn't allow groups of size 0 you would obviously get $C(69,4)$ solutions.

But since you want non-negative solutions, you have to do something a bit more subtle: say that you take a solution to your problem and add 1 to each of $a,b,c,d$ and $e$, then you get a strictly positive solution to $a+b+c+d+e=75$ (and you should convince yourself that a positive solution to this equation yields a non-negative solution to your original problem). But clearly, for our discussion in the previous paragraph, there are $C(74,4)$ solutions to that problem!

David Kohler
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This problem belongs to "$k$-combination with repetitions" problems. The general formula is like this: $X_1+X_2+X_3+X_4+\cdots+X_n=r$. Can you see the problem in your answer? According to this, $r=70$ and $n=5$ so the combination should be $$C(n+k-1,n-1) = C(74,4) = C(n+k-1,r) = C(74,70) = 1,150,626.$$

Cookie
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Rakibul Islam
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