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What is the number of solutions to $(x_1+x_2+x_3)(x_4+x_5+x_6+x_7)=77$, where $x_1,x_2,\dots,x_6,x_7$ are non negative integers? Obviously the $2$ multipliers are $1 \times 77$ and $7 \times 11$, so we have $x_1+x_2+x_3 = 1,7,11,77$ and $x_4+x_5+x_6+x_7 = 1,7,11,77$.

RobPratt
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Avi Tal
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  • Have you heard of stars and bars? – octave Mar 22 '21 at 22:25
  • Obviously I am not an expert... – Avi Tal Mar 22 '21 at 22:26
  • You should reverse one of $1,7,11,77$. – user Mar 22 '21 at 22:28
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    That's alright. It turns out that the number of nonnegative integer solutions to $x_1+x_2+\cdots+x_k = n$ is $\binom{n+k-1}{n}$, where $\binom{}{}$ is the binomial coefficient. A reference can be found at https://math.stackexchange.com/questions/910809/how-to-use-stars-and-bars-for-solving-number-of-integral-solutions-to-an-equalit. For your problem, you can handle the cases separately: the number of ways for $x_1 + x_2 + x_3 = 1$ and $x_4 + x_5 + x_6 + x_7 = 77$, etc. – octave Mar 22 '21 at 22:30
  • @octave - Would you like to post this as an answer? – Avi Tal Mar 22 '21 at 22:37

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Recall the method of "stars and bars," which tells us that the number of nonnegative integer solutions to the equation $$ x_1 + x_2 + \cdots + x_k = n $$ is $$ \binom{n+k-1}{n}. $$

There are four cases, as you pointed out. The tuple $(x_1 + x_2 + x_3, x_4 + x_5 + x_6 + x_7)$ can only equal $(1, 77)$, $(7, 11)$, $(11, 7)$, or $(77, 1)$. In the first case, where $x_1 + x_2 + x_3 = 1$ and $x_4+x_5+x_6+x_7=77$, the number of solutions is $$ \binom{3}{1}\binom{80}{77}. $$

Applying this reasoning for the other three cases, we get that the total number of solutions is $$ \binom{3}{1}\binom{80}{77} + \binom{9}{7}\binom{14}{11} + \binom{13}{11}\binom{10}{7} + \binom{79}{77}\binom{4}{1}. $$

octave
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  • Thanks allot for your answer! And how about this: |x|+|y|+|z| = 30? Without the absolute value I get 32 choose 30 (496). With the absolutes my intuition was to multiply this by 8, but I actually get 3602 and I don't know why... – Avi Tal Mar 23 '21 at 05:06
  • You can ask a separate question (not in these comments) if it’s a different equation. – octave Mar 23 '21 at 09:54