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what is the number of integer solutions to $$x_1+x_2+x_3+x_4+x_5=18$$ with $$x_1\ge1\;\;\;x_2\ge2\;\;\;x_3\ge3\;\;\;x_4\ge4\;\;\; x_5\ge5$$

I know I have to use this formula $$\frac{(n+r-1)!}{(n-1)!\;r!}= {{n+r-1}\choose r}$$

My instinct says that I should use $n=18-1-2-3-4=18-15=3$ and $r=5$ but I m not sure it makes sense?

Anyone can help me please?

Davide Morgante
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voldetort
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    You can use standard methods if you replace $x_i$ with $y_i=x_i-i$. Then, since $1+2+3+4+5=15$ we just have $\sum y_i=3$ and $y_i≥0$. – lulu Aug 28 '18 at 18:26
  • Well ok but can you explain it a bit better? what is the standard method I should use and how does it help me get to a conclusion – voldetort Aug 28 '18 at 18:28
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    The standard method goes by the name Stars and Bars. Though, really, it isn't that hard to count the number of $5$-tuples that add to $3$. – lulu Aug 28 '18 at 18:29
  • I am asking because I haven t studied this method and I m trying to figure it out on my own. Also it can be a bit difficult to count the number of 5-tuples if the number they have to add to is greater than 3 so I thought it would be useful to know the method in general. Thanks anyway ^^ – voldetort Aug 28 '18 at 18:36
  • The link I provided gives closed formulas (plus proofs) of the relevant formulas. – lulu Aug 28 '18 at 18:37

2 Answers2

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We can simply give the required value to each pirate before counting. We don’t have any choice for them, so there is exactly one way to do this. Then, we distribute the remaining $18-5-4-3-2-1=3$ bars among the pirates.

This gives $$\dbinom{3+5-1}{3}=\dbinom{7}{3}$$

Edit:

Given $$x_1+x_2+x_3+x_4+x_5=18$$and $$x_1\ge1\;\;\;x_2\ge2\;\;\;x_3\ge3\;\;\;x_4\ge4\;\;\; x_5\ge5$$ Let $$y_1 = x_1 − 1, y_2 = x_2 − 2, y_3 = x_3 − 3, y_4 = x_4 − 4, y_5 = x_5 − 5$$ $$x_1+x_2+x_3+x_4+x_5=18$$$$(y_1 + 1) + (y_2 + 2) + (y_3 + 3) + (y_4 + 4) + (y_5 + 5) =18$$$$y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 3$$ There are $\dbinom{3+5-1}{3}=35$ solutions

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We can change the problem from its original form $$x_1+x_2+x_3+x_4+x_5=18$$ $$x_1\ge1,x_2\ge2,x_3\ge3,x_4\ge4,x_5\ge5$$ to $$y_1+y_2+y_3+y_4+y_5=(x_1-1)+(x_2-2)+(x_3-3)+(x_4-4)+(x_5-5)=3$$ $$y_1\ge0,y_2\ge0,y_3\ge0,y_4\ge0,y_5\ge0$$ and use generating functions approach, giving the following function $$(1+y+y^2+...+y^k+...)^5$$ and the coefficient of $y^3$ term is the answer, which is 35. Here is another example to look at (which contains another link to another example ...).

rtybase
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