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Find the number of non-negative integers to
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 57$$ where $$x_1 \lt 3, x_3 \ge 4$$ I first found the total number of solutions N=$57+6-1\choose57$=6471002, then the total number of equations for the restriction $x_1 \ge 3$, N(P1)=$54+6-1\choose54$=3425422, then for the restriction $x_3 \le 3$, N(P2)=$4+6-1\choose4$=126 and I have to find the number of solutions for both the restrictions N(P1P2), but I don't know the number I should plug in the combination formula or if my calculations are correct.

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  • $\begingroup$ The way you handled the $x_1\geq 3$ restriction you can also handle the $x_3\geq 3$ restriction. Which is to say, remove the restriction and say you want $x_1+x_2+x_3+x_4+x_5+x_6 = 54$ instead. That's one less thing to worry about. $\endgroup$
    – Arthur
    Jan 14, 2020 at 21:39
  • $\begingroup$ Can you elaborate? $\endgroup$
    – lina
    Jan 14, 2020 at 21:41
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    $\begingroup$ How did you get $54+6-1\choose54$? That's really all there is to it. Just do it to $x_3$ rather than $x_1$, and simplify the problem before trying to use inclusion-exclusion. $\endgroup$
    – Arthur
    Jan 14, 2020 at 21:43
  • $\begingroup$ I see now that we have $x_3\geq 4$ rather than $x_3 \geq 3$. So apologies for the confusion there. Still, that's something I would suggest you take care of first, and then set to find the number of solutions. Like was suggested in the naswer below. $\endgroup$
    – Arthur
    Jan 14, 2020 at 22:22

1 Answer 1

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This problem is equivalent to distribute $57$ candies to $6$ kids. To eliminate $x_3\ge4$ restriction, let's give $4$ candies to $x_3$. So problem is now

$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 53$ with $x_1\le2$

I prefer Stars and Bars method see here.

$x_1=0:\quad$ $ x_2 + x_3 + x_4 + x_5 + x_6 = 53$. This implies we have $53$ stars and $4$ bars and we use Permutations with Repetition see here

$$\frac{57!}{53!4!}$$

$x_1=1:\quad$ $ x_2 + x_3 + x_4 + x_5 + x_6 = 52$ $$\frac{56!}{52!4!}$$

$x_1=2:\quad$ $ x_2 + x_3 + x_4 + x_5 + x_6 = 51$

$$\frac{55!}{51!4!}$$

So the answer is $$\frac{57!}{53!4!}+\frac{56!}{52!4!}+\frac{55!}{51!4!}$$

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  • $\begingroup$ How did you get x1≤2? $\endgroup$
    – lina
    Jan 14, 2020 at 21:52
  • $\begingroup$ Because $x_1<3$ and it is non-negative integer, $x_1<3\equiv 0\le x_1\le2$ $\endgroup$
    – OkkesDulgerci
    Jan 14, 2020 at 21:56

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