no. of ways

[fallow pewter]

let $n{1}, n{2}, n{3}, n{4}, n{5}$ be positive integers such that $n{1} < n{2} < n{3} < n{4} < n{5}$ and $n{1} + n{2} + n{3} + n{4} + n{5} = 20$, then find the no. of such $n{x}$

[soft matrix]

Sypse

[arctic lodge]

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[fallow pewter]

all I tried was getting all the possible values without the inequality and then subtracting those cases where 2/3/4/all of them were equal, but things were getting pretty messy
so Idk where to go now

[void minnow]

Hm... Looks like something stars and bars can deal with.
https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
Not sure how to deal with the first constraint, though.

[fallow pewter]

I did that only and then the issue starts, I think it can only be done manually

[fair stag]

"The no. Of such n_x"? Can we see a screenshot of the original question?

[fallow pewter]

I dont have it currently, but we just need the number of ordered pairs of (n1,n2,n3,n4,n5)

[void minnow]

Hm... I suppose since n1, ..., n5 must be different positive integers, then n1 must be at least 1, n2 must be at least 2, etc. So, if we take k(i) = n(i) - i, then we get an equation k1 + ... + k5 = 10, where k1, ..., k5 are nonnegative integers and k1 ≤ k2 ≤ ... ≤ k5.
Not sure if this makes the problem easier, though...

[fair stag]

It actually makes it harder. In fact, what you did is the exact inverse of what we do to count multisets.

[void minnow]

Oh, I see...

[rigid gyro]

There’s a specific formula for the amount of solutions to x1+x2…xk=n, for non-negative x_i, which is (n+k-1)C(k-1) (this is the thing LordDarpinger referenced)

[fallow pewter]

ik about it

[rigid gyro]

What I’m wondering is if these ordered pairs have to be distinct

[fallow pewter]

they would need to because of the inequality no?

and those no. of distinct ways are the answer

because order wont matter

[rigid gyro]

Yes

You can assume that for every ordered pair, you can arrange it to satisfy the inequality

[fallow pewter]

hm so ig manually doing it is the only way right?

[rigid gyro]

I suppose that you have to find the amount of pairs that do not satisfy the inequality

[fallow pewter]

which will be a bigger headache

[rigid gyro]

I have an idea as to how to do it without manual calculation

[fair stag]

I also have an idea.

[rigid gyro]

No, do it for only one case

Then generalize it

[fallow pewter]

and finding the ones with the inequality will be faster because its only 7

[fair stag]

Actually, @void minnow, I just realized your method will help.

[vital phoenix]

Use stars and bars, remove cases in which all 5 are not distinct, then divide by 5!

[flint cloud]

this makes the question very easy to do manually

[forest jasper]

@fallow pewter

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[fallow pewter]

+close

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