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How many solutions are there to the equation below, if $x_i$ is a positive integer > 1: $$\sum_{1}^{6} x_i = 29$$

I also have to do this for x1<=5, however I imagine that's a similar process.

So, my question is based on $x_i>1$. Where is a good starting point?

Edit: I feel as though I should use bars and stars in some way. We have 6 bars and 17 stars, then multiply that by the possible number of ways that we can order the bars (i.e. $x_1,x_2,x_3,x_4,x_5,x_6 \space \text{vs} \space x_1,x_2,x_3,x_4,x_5,x_6$)?

After realizing I can have a min of 2 and max of 19 what's a step in the right direction?

Also, I just realized I've had the idea in my mind that 2 + 2 + 2 + 2 + 2 + 19 and 19 + 2 + 2 + 2 + 2 + 2 would be different solutions, but I now believe those would count as the same. Just as some clarification, plus to explain why I wanted to multiply by 6! in my previous attempt

Starlight
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user143918
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  • That's correct; sorry I should have mentioned that, which means the max is 19, I believe. I feel as though I should use bars and stars in some way since that's a big part of what we're learning, would one way to go about it be to say we have 6 bars and 17 stars, use bars and stars then multiply that by the possible number of ways that we can order the bars (i.e. x1,x3,x2,x4,x5,x6 vs x1,x2,x3,x4,x5,x6)? – user143918 Oct 22 '14 at 03:38
  • Scratch that; it doesn't work, at least how I'm trying it. What's a good step after realizing the max is 19 and the min is 2? – user143918 Oct 22 '14 at 03:42
  • There's no need to put [Answered] in the title. Accepting an answer is enough, and you've done that. – Asaf Karagila Oct 22 '14 at 04:49

3 Answers3

1

The generating function is:

$$f(x)=(x^2+x^3+x^4\cdots)^6=x^{12}(1+x+x^2+x^3+\cdots)^6=x^{12}\frac{1}{(1-x)^6}=x^{12}\sum_{i=0}^{\infty}\binom{6+i-1}{i}x^i$$

then:

$$\binom{6+17-1}{17}=\binom{22}{17}=26334$$

AsdrubalBeltran
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1

We can also use the Stars and bars argument -- associate with each $x_i$ the variable $y_i$ such that $x_i = 2 + y_i$. Then, we have an equivalent problem in:

$$\sum_{i=0}^6 y_i = 17$$

(subject to $y_i \ge 0$). Using the stars and bars argument we immediately have $\binom{22}{5}$ as our answer.

Irvan
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0

I solve a simpler problem. Assume x1=x2=x3=2

Then there are 9 solutions with x4=2.

There are 8 solutions with x4=3

There are 6 solutions with x4=4

There are 5 solutions with x4=5

There are 3 solutions with x4=6

There are 2 solutions with x4=7

That is a sum of 33 solutions, and we must multiply by 6!=720 because I assume answers are the same up to rearranging of indices.

However I have ignored many solutions.

user145600
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