Vtyjkyuk

what is the number of integer solutions to
$$x_1+x_2+x_3+x_4+x_5=18$$ with $$x_1ge1;;;x_2ge2;;;x_3ge3;;;x_4ge4;;; x_5ge5$$

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

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?

We can change the problem from its original form
$$x_1+x_2+x_3+x_4+x_5=18$$
$$x_1ge1,x_2ge2,x_3ge3,x_4ge4,x_5ge5$$
to
$$y_1+y_2+y_3+y_4+y_5=(x_1-1)+(x_2-2)+(x_3-3)+(x_4-4)+(x_5-5)=3$$
$$y_1ge0,y_2ge0,y_3ge0,y_4ge0,y_5ge0$$
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 ...).

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 $$dbinom3+5-13=dbinom73$$

Edit:

Given $$x_1+x_2+x_3+x_4+x_5=18$$and $$x_1ge1;;;x_2ge2;;;x_3ge3;;;x_4ge4;;; x_5ge5$$
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 $dbinom3+5-13=35$ solutions