Thread: 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

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 the answer to this question is the coefficient to the term.

Luckily the is an online resource to do just that.

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.

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.

.

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)
AD=(11C3)
BC=(11C3)
BD=(11C3)
CD=(11C3)
ABC=(3C3)
ABD=(3C3)
ACD=(3C3)
BCD=(3C3)

|ABCD|=|U|-|A|-|B| -|C|-|D|+|AB|+|AC|+|BC|+|AD|+|BC|+|BD|-|ABC|-|ABD|-|ACD|-|BCD|

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

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)
AD=(11C3)
BC=(11C3)
BD=(11C3)
CD=(11C3)
ABC=(3C3)
ABD=(3C3)
ACD=(3C3)
BCD=(3C3)

|ABCD|=|U|-|A|-|B| -|C|-|D|+|AB|+|AC|+|BC|+|AD|+|BC|+|BD|-|ABC|-|ABD|-|ACD|-|BCD|

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

That is great!
I hope you learned something.
You really do need to study generating functions.

You really do need to study generating functions.

As soon as I have an exam that requires me to know it