[Discrete Maths] Counting ,Number of solutions to the equation

Question

How many solutions are there to the equation
x1 + x2 + x3 + x4 + x5 = 21,
where xi , i = 1, 2, 3, 4, 5, is a nonnegative integer such that:

0 ≤ x1 ≤ 3 , 1 ≤ x2 < 4  and 15≤x3 ?

Comments

I tried moving lower constraint from x2(1) and x3(15) and RHS becomes 21-6=5.

Then i find the coefficient of x^5 which is coming as -106 but answer in book is 106,Does sign matters here?I know number of solutions will be +ve only but if it comes -ve then can we ignore sign or do we give 0 as answer?

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$\binom{24}{4}$ - $\binom{8}{4}$ - $\binom{21}{4}$ + $\binom{5}{4}$ - $\binom{20}{4}$ + $\binom{4}{4}$ + $\binom{17}{4}$   ????

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Given answer is 106.

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$\binom{9}{4}$  - $\binom{6}{4}$ - $\binom{5}{4}$  = 106.

Sorry by mistake in the previous one I considered  x3 <= 15.

Answer 1

Hope this helps.

Comments

I started with the same approach. I have written all the terms in power of x.And then i find coefficient of x^5 .And it came as -106.So we can ignore the coefficient and select 106 as the answer?

If you didn't understand my approach,i will upload an image.let me know

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How y2 = x2-1?

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Given that x2>=1. So by adding -1 on both sides of inequality:

(x2-1)>=(1-1) => (x2-1)>=0.

Now instead of writing x2-1 everywhere further, replace it with y2 such that y2=x2-1 and y2>=0.

Answer 2

answer: -106