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What is the number of non negative integral solutions of $5w+3x+y+z=100$, such that $w≤x≤y≤z$.

I have the answer key to this but I do not understand how to even start. The second inequality condition is what is troubling me the most. Are there specific ways to approach this?

Edit: I have a solution of this but I am unable to make any sense of it.

Let a = w, b = x - w, c = y- x, and d = z - y. Observe that a,b,c,d are all nonnegative integers, since w, x, y, z are nonnegative. Then the number of solutions to the given equation is equivalent to the number of solutions to the equation 5a + 3(a + b) + (a + b + c) + (a +b + c+ d) = 10a + 5b + 2c + d = 100. Next, we compute the total number of combinations by considering the sum in increments of 10, where there are a total of 100/10 = 10 increments to consider. There are 3 possible cases:

(A) The only increments of 10 are 10, 5 + 5, 2 + 2 + 2 + 2 + 2, and 1 + 1 + …+ 1 =10. The number of solutions is equivalent to computing the number of ways of placing 10 balls in 4 urns, or of placing 3 dividers in 10 increments. Hence, there are 286 possible solutions.

(B) In addition to the above 4 increments, there is 1 increment of 10 consisting of some combination of 5s, 2s, and Is. In total, there are 7 ways of achieving this: 2, 2 +2, 2 + 2 + 2, 2 + 2 + 2 + 2, 5, 5 + 2, and 5 + 2 + 2 (with each sum padded with is to equal 10). The number of ways of placing the 4 even increments of 10, 5.2, 2.5, and 1.10 is equivalent to the number of ways of placing 3 dividers in 9 increments. Hence, there are 1540 possible solutions.

(C) In addition to the above 4 increments, there are 2 increments of 10 consisting of some combination of 5s, 2s, and Is. In total, there are 2 ways of achieving this: 5 + 2 + 2 + 2 and 5 + 2 + 2 + 2 + 2 (with each sum padded with Is to equal 10). The number of ways of placing the 4 even increments is equivalent to the number of ways of placing 3 dividers in 8 increments. Hence, there are 330 possible solutions. In total, there are 286 + 1540 + 330 = 2156 total solutions to the original equation.

A shubh
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  • I would combine the two conditions to build other inequalities. For instance, we have $$ 10w \le 5w + 3x + y + z \le 10 z $$ which implies $w \le 10$ and $z \ge 10$. – dvdgrgrtt Feb 17 '24 at 07:28
  • After making the appropriate substitutions you might like to check out the approach offered by Mike Earnest in this question. – Sahaj Feb 19 '24 at 05:44

1 Answers1

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Introduce non-negative integers $a,b,c$ as auxiliary variables

Put $\;x = w+a,\;y = w+a+b,\; z=w+a+b+c,\;$ to get

${5w + 3(w+a) + (w+a+b) +(w+a+b+c)}$

$\rightarrow 10w + 5a +2b+c =100$

Now you can use generating functions to solve it

The answer will be the coefficient of $x^{100}\;$ in expansion of $\frac{1}{1-x^{10}}\cdot\frac{1}{1-x^5}\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x}$

giving answer $= 2156$

true blue anil
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