<s>[INST] How many different systems are there? A Football team has 10 players besides the Keeper, who are allocated into 3 positions, the defense, the center and the attack. A system is the allocation of the players into these positions.
For example the system 4 4 2 means that there are 4 players at the defense, 4 players at the center and 2 players at the attack.
There are 5 players available for the defense, 6 players for the center and 4 players for the attack.
How many different systems can be formed, when we know that at each position must be at least one player??

I found the following sentence:
$$ \text{ The number of integer solutions of the equation}  x_1+x_2+ \dots + x_n=r , x_1>a_1 , x_2>a_2 , \dots, x_n>a_n \text{ can be found with the formula } \displaystyle{\frac{r-a_1-a_2- \dots -a_n-1}{n-1}} $$
Can I do it as followed?
$$x_1+x_2+x_3=10 , x_1 \leq 5, x_2 \leq 6, x_3 \leq 4, x_1,x_2,x_3 \geq 1 $$
$$y_1=x_1-1, 0 \leq y_1 \leq 4 $$
$$y_2=x_2-1, 0 \leq y_2 \leq 5 $$
$$y_3=x_3-1, 0 \leq y_3 \leq 3 $$
$$ y_1+1+y_2+1+y_3+1=10 \Rightarrow y_1+y_2+y_3=7 , y_1,y_2,y_3 \geq 0$$
There are $\binom{3+7-1}{7}=36$ integer solutions of the equation $  y_1+y_2+y_3=7 , y_1,y_2,y_3 \geq 0$
We have to substract the cases:


*

*$x_1>5: \binom{10-5-1}{2}=6 $

*$x_2>6: \binom{10-6-1}{2}=3$

*$x_3>4: \binom{10-4-1}{2}=10 $
So,there are $36-6-3-10=17$  different systems that can be formed.
Or couldn't we do it this way?
 [/INST] i have another idea. I used a generating function. Here it is $(x+x^2+x^3+x^4+x^5+x^6)*(x+x^2+x^3+x^4+x^5)*(x+x^2+x^3+x^4)$. If you expand this expression you can read off the number of systems can be formed. The Expansion is $x^3+3 x^4+6 x^5+10 x^6+14 x^7+17 x^8+18 x^9+17 x^{10}+14 x^{11}+10 x^{12}+6 x^{13}+3 x^{14}+x^{15}$
The exponents are the number of team members. In your case 10. The related coefficient is the number of systems can be formed.
greetings,
calculus
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