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How many nonnegative integer solutions are there to the equation $$A_1 + A_2 + A_3 + A_4 = 25$$ such that $A_1 \leq 12$ and $A_4 \leq 10$?

I am completely lost on how to solve such a question.
If someone could give me a detailed solution, I would greatly appreciate it.

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    $\begingroup$ I feel that there's a mistake in the phrasing of the question - what's $A_5$? $\endgroup$ Dec 6, 2018 at 6:25
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    $\begingroup$ Sorry about that, its fixed now $\endgroup$
    – Steve
    Dec 6, 2018 at 6:29
  • $\begingroup$ How many choices are their for A1? 12 (13 if you include 0 I suppose). How many choices are there for A4? 10? So how many pairs of A1 and A4 can you have? $\endgroup$ Dec 6, 2018 at 6:36
  • $\begingroup$ I'm not quite sure....The question was stated exactly as above. $\endgroup$
    – Steve
    Dec 6, 2018 at 6:40
  • $\begingroup$ @DanielOnMSE More is needed since have to assign $A_2,A_3.$ $\endgroup$
    – coffeemath
    Dec 6, 2018 at 7:04

2 Answers 2

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We can think of it in this way:

We find all cases for the equation without restriction of $A_1\leq12$ and $A_4\leq10$. Then we fill in these cases with a 2-circle Venn Diagram, the first circle ($P$) with cases satisfying $A_1>12$ (yes, $>$), and the second circle ($Q$) with cases satisfying $A_4>10$. Then what we want to find is $P'\cap Q'$ i.e. $\xi - P - Q + P\cap Q$.

Now we have to solve for

$$A_1 + A_2 + A_3 + A_4 = 25\ \fbox*$$ when $\fbox1\ A_1>12\ (A_1\geq13)$, when $\fbox2\ A_4>10\ (A_4\geq10)$, and when $\fbox3$ both occurs.

From now, we think of those 4 "$A$"s as bins, and numbers as "balls".

For $\fbox*$

Using stars and bars we get ${28\choose3}=3276$

Case $\fbox1$

Then it's easy to eliminate the restriction. We fill in box $A_1$ with 13 balls, and then we get an equation we need to solve for (with no restrictions!): $$A_1 + A_2 + A_3 + A_4 = 12$$

Using stars and bars the answer is ${15\choose3}=455$

Case $\fbox2$

Similarly we get $$A_1 + A_2 + A_3 + A_4 = 14$$ and hence we get ${17\choose3}=680$

Case $\fbox3$

Similarly we get $$A_1 + A_2 + A_3 + A_4 = 1$$ and hence we get ${4\choose3}=4$

Hence we have $3276-455-680+4=2145$ which is what we want.$\ \square$

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  • $\begingroup$ I understand your process, but i'm a bit confused about why you can't make a substitution for A1 and A2 after you make A1≥13 and A≥11 $\endgroup$
    – Steve
    Dec 6, 2018 at 16:16
  • $\begingroup$ yeah I can make a substitution of $x=A_1-13$ and $y=A_4-11$ (the process is similar) $\endgroup$ Dec 7, 2018 at 11:42
  • $\begingroup$ May I know where did you get this question (which topic are you studying) so I can give you a proof related to the topic? :) $\endgroup$ Dec 7, 2018 at 11:43
  • $\begingroup$ It was from a past midterm so i'm not exactly sure. But i've found that using generating functions to solve these problems are easier. $\endgroup$
    – Steve
    Dec 7, 2018 at 19:07
  • $\begingroup$ Well you can use generating functions by expanding $(\sum_{k=0}^{12}x^k)(\sum_{k=0}^{25}x^k)^2(\sum_{k=0}^{10}x^k)$ (I suggest doing it with WolframAlpha) $\endgroup$ Dec 8, 2018 at 3:49
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Sum for $A_1$ from $0$ to $12$ aand $A_4$ from $0$ to $10$ of $25-A_1-A_4+1$ where the last part is from stars and bars for the possible values of $A_2,A_3.$ [There's only one bar.] If I entered it right it's $2145.$

It's lucky that $12+10=24\le 25$ here. [edit-- lucky that $12+10=22 \le 25.$]

More explanation: You can have any of the possible values up to $12$ for $A_1$ and up to $10$ for $A_4$ (including $0$ for each.) These are distinct possibilities for assigning two of the four letters. Once $A_1$ and $A_4$ have been assigned, the overall sum of $25$ needs $25-A_1-A_4$ more to go, between $A_2$ and $A_3$, and this can be completed in $[25-A_1-A_4+1]$ ways, hence that is the value summed.

One more thing-- I just put this sum in a calculator to get the number, but it may be possible to do it formally using sum formulas applied to parts of the thing summed.

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  • $\begingroup$ yea that answer matches the answer key. But would it possible for you to make your solution more explicit? $\endgroup$
    – Steve
    Dec 6, 2018 at 7:27
  • $\begingroup$ @Steve Just added some explanation at end. $\endgroup$
    – coffeemath
    Dec 6, 2018 at 7:55

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