# Thread: Find the number of integral solutions

1. ## Find the number of integral solutions

Find the number of integral solutions of x1 + x2 + x3 + x4 = 28

where each xi ≥ 1, and in addition each xi ≤ 8. Write your answer in terms of binomial coefficients.
I am a little confused by this question.

My solution would be to take the number of solutions where xi ≥ 1. Then I would subtract from that the total number of solutions where xi ≥9. This would leave the number of solutions where xi ≥ 1, and in addition each xi ≤ 8.

But when I try to find solutions where xi ≥9 I realized this is not possible.

Could someone give me a nudge in the right direction (without giving the answer).

Thanks

2. ## Re: Find the number of integral solutions

Find the number of integral solutions of
x1 + x2 + x3 + x4 = 28
where each xi ≥ 1, and in addition each xi ≤ 8. Write your answer in terms of binomial coefficients.
This is a nightmare of a question if one does by cases.
It is easy if you use generating functions.
Expand $\left( {\sum\limits_{k = 1}^8 {x^k } } \right)^4$ the answer to this question is the coefficient to the $x^{28}$ term.

Luckily the is an online resource to do just that.

3. ## Re: Find the number of integral solutions

Ya but this could show up on the exam so I really want to be able to understand and solve it.

I am not quite sure what generating functions are, and I know they were not covered in class, so there must be another way to solve it that I could understand :P.

It says put your answer in terms of binomial coefficients, so i am sure that it is expected it solved this way.

4. ## Re: Find the number of integral solutions

Originally Posted by ehpoc
Ya but this could show up on the exam so I really want to be able to understand and solve it. It says put your answer in terms of binomial coefficients, so i am sure that it is expected it solved this way.
Well ok. But don't ask me to explain.
I will tell you that it comes from extended applications of inclusion/exclusion.

$\sum\limits_{k = 0}^3 {( - 1)^k \binom{27-8\cdot k}{3}\binom{4}{k}}=35$.

5. ## Re: Find the number of integral solutions

I think I solved it!

U=(27C3)
A=(19C3) =total ways with only x1>8
B=(19C3) =total ways with only x2>8
C=(19C3) =total ways with only x3>8
D=(19C3) =total ways with only x4>8
AB=(11C3)
AC=(11C3)
BC=(11C3)
BD=(11C3)
CD=(11C3)
ABC=(3C3)
ABD=(3C3)
ACD=(3C3)
BCD=(3C3)

=(27C3)-4(19C3)+6(11C3)-4=35

6. ## Re: Find the number of integral solutions

Originally Posted by ehpoc
I think I solved it!

U=(27C3)
A=(19C3) =total ways with only x1>8
B=(19C3) =total ways with only x2>8
C=(19C3) =total ways with only x3>8
D=(19C3) =total ways with only x4>8
AB=(11C3)
AC=(11C3)
BC=(11C3)
BD=(11C3)
CD=(11C3)
ABC=(3C3)
ABD=(3C3)
ACD=(3C3)
BCD=(3C3)