Im so confused.

[cinder topaz]

How many non-negative integer solutions are there to:
𝑥 + 𝑦 + 𝑧 = 12
such that 𝑥 ≤ 5, 𝑦 ≤ 6, and 𝑧 ≤ 7?

I think this is meant to be done using inclusion exclusion principle?

[storm vigil]

Yes

Define X to be the set of solutions with x>5
Y be the set with y>6 and Z be that of z>7

Then you basically just find
Total solutions - |X|-|Y|-|Z|+|X and Y|+|Y and Z|+|Z and X|-|X and Y and Z|

You can calculate them individually by using auxiliary variables
For example for |X and Y|
I would define x’=x-6,y’=y-7
And use stars and bars on that

[cinder topaz]

I dont get it. Can you solve it?

@storm vigil

[storm vigil]

Do you know what stars and bars is

Ok

Let’s say x y and z didn’t have the upper bound restriction

Then how would you solve it

[cinder topaz]

Not really. Is that

Combination of n+k-1
K-1?

[storm vigil]

That’s the formula

It’s the number of ordered pairs of nonnegative integers that solve x1+x2+…xk = n

[cinder topaz]

So again how would I solve this

Why would you take x’=x-6 and y’=y-7

[storm vigil]

Right

So you want to find the number of solutions to x +y+z where all three are nonnegative
And subtract all the cases where at least one of them is greater than their upper bound

If I’m trying to find the number of solutions to x+y+z=12 where x is at least 6 and y is at least 7

If I define x’=x-6 and y’=y-7

The equation becomes x’+y’+z=-1

Where all three are now ag least 0

Ok so this one has no solutions

You can try solving for |X| ig then

And |Y| and |Z|

Since the higher order terms vanish

[random carbon]

Try to divide this qn into cases

1st put z = 7

To solve number of ways of x + y = 5 with their respective constraints

And go on

Their will be total of 7 cases

It's easy way to solve these kinda qns

[wet cosmos]

Are you kidding me?

[round falcon]

wjats tje kidding part