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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?

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Could you use some kind of spell checker and try to review posts before submitting? –  Kaster Apr 28 at 7:07

2 Answers 2

up vote 2 down vote accepted

How about enumerating all possibilities?

Starting with 1 defender (the minimum) and 4 attackers (the maximum), we get:

  1. 1-5-4
  2. 1-6-3
  3. 2-4-4
  4. 2-5-3
  5. 2-6-2
  6. 3-3-4
  7. 3-4-3
  8. 3-5-2
  9. 3-6-1
  10. 4-2-4
  11. 4-3-3
  12. 4-4-2
  13. 4-5-1
  14. 5-1-4
  15. 5-2-3
  16. 5-3-2
  17. 5-4-1

Hmm. That is not what you have. ;)

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2  
The only alternative I'm aware of are those generating functions. Since I suspect you haven't learned about them (yet), I think you are supposed to enumerate them. –  I like Serena Apr 28 at 21:00
1  
Generally where you're supposed to count something which can be A OR B AND C etcetera, and where the total is some fixed amount. In your case, where the defenders can be 1 OR 2 OR 3 OR 4, AND where the center can be 1 OR 2 OR... All that with a total of 10. This is represented by $(x + x^2 + x^3 + x^4) \cdot (x + ...)$. –  I like Serena Apr 28 at 23:11
2  
$$x_1+x_2+x_3=10 , x_1 \leq 5, x_2 \leq 6, x_3 \leq 4, x_1,x_2,x_3 \geq 1 $$ $$y_1=x_1-1, 0 \leq y_1 \leq 4 $$ $$y_2=x_2-1, 0 \leq y_2 \leq 5 $$ $$y_3=x_3-1, 0 \leq y_3 \leq 3 $$ $$ y_1+1+y_2+1+y_3+1=10 \Rightarrow y_1+y_2+y_3=7 , y_1,y_2,y_3 \geq 0$$ 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? –  Mary Star May 3 at 21:55
1  
This appears to work! Mostly because apparently there is no overlap. $\ddot \smile$ I do not understand the formula you gave in your previous comment though. It seems to me that it is not correct. –  I like Serena May 5 at 21:54
1  
I thought about the exercise again, since the formula was wrong..Here's my idea: $$D+C+A=10,1 \leq D \leq 5,1 \leq C \leq 6,1 \leq E \leq 4$$ $$D'=D-1 \geq 0$$ $$C'=C-1 \geq 0 $$ $$ A'=A-1 \geq 0$$ $$D'+C'+A'=10-(1+1+1) \Rightarrow $$ $$D'+C'+A'=7, \text{ where }D', C',A' \geq 0 :$$ $$\binom{3+7-1}{3-1}=\binom{9}{2}=36$$ These are the cases that at each line there is at least one player. Then we have to substract the following: –  Mary Star May 18 at 14:10

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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I got stuck right now :/ could explain me why you multiplied these specific functions?? –  Mary Star Apr 28 at 9:44
1  
I cannot tell here the whole theory beyond generating funtion. This link might help you: [1]: mathdb.org/notes_download/elementary/algebra/ae_A11.pdf (Chapter A, p.9) Here the defence, center and attack have at least one player. So there is no $x^0=1$ in each bracket. In the team there are not more than 5(d), 6(c) and 4(a) player. That are the limits for the exponents. –  calculus Apr 28 at 10:12
    
Ok, I will think about your solution... Is my way wrong? –  Mary Star Apr 28 at 12:02
    
the result is wrong. And your transformation doesn´t really help. You have also to consider the bounderies of the $z_i$´s –  calculus Apr 28 at 13:02
    
I edited my first post... Could you tell me if it is correct now? –  Mary Star May 4 at 16:16

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