# Find the number of non-negative integer solutions of the equation $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 57$ where $x_1 \lt 3, x_3 \ge 4$

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.

• 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. Jan 14, 2020 at 21:39
• Can you elaborate?
– lina
Jan 14, 2020 at 21:41
• 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. Jan 14, 2020 at 21:43
• 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. Jan 14, 2020 at 22:22

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 $$\frac{57!}{53!4!}+\frac{56!}{52!4!}+\frac{55!}{51!4!}$$

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