3
  1. $x_1 + x_2 + x_3 + x_4 + x_5 = 100$
  2. $x_1 + x_2 + x_3 + x_4 + x_5 \leq 100$
  3. $x_1 + x_2 + x_3 + x_4 + x_5 < 100$ with all $x_i > 0$

For the first one I said that the answer was $104C4$ and for the second one I said the answer was $105C5$. Are these two correct? If not, how do I do it? As for question 3, I have no idea how to do it. Any help would be appreciated.

N. F. Taussig
  • 76,571
John1234
  • 31
  • 1
  • 1
  • 2
  • Your answer will depend on whether you count $x_1 = 40, x_2 = 60$ (and the rest zero) as distinct from $x_1 = 60, x_2 = 40$ (and the rest zero). Which one are you assuming? – Michael Seifert Jan 25 '16 at 16:00
  • This is how I did it: (n+r-1)C(r-1) and got the answers. Is this wrong? – John1234 Jan 25 '16 at 16:04
  • Those would be correct if you are counting the two options I listed as distinct. See my answer below. – Michael Seifert Jan 25 '16 at 16:06
  • What about question 3? It has says that xi > 0 and I am not sure what to do. I tried doing it this way: I added another variable x6+1, such that x1+x2+x3+x3+x4+x5+x6=99. Then I used: (n-1)C(r-1). Is this correct way of doing it? 98C5 – John1234 Jan 25 '16 at 16:12
  • Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Jan 25 '16 at 16:16

2 Answers2

5

The first two are correct, assuming you are counting things like $x_1 = 60, x_2 = 40$ as distinct from $x_1 = 40, x_2 = 60$. The idea here is that you're spreading a certain number of balls into a certain number of bins. The equation you used is usually derived via a stars and bars argument.

For part (a), you're basically distributing 100 balls into 5 bins, so the solution would be $104 \choose 4$. For part (b), you can imagine having 6 bins instead; the 100 balls are distributed between them, and the number of balls in the first five is then less than or equal to 100. The answer is therefore $105 \choose 5$.

Since this looks like a homework problem, I won't give you the straight answer to part (c), but think of it this way: if I handed you 100 balls and told you to put them into 5 bins such that there's at least one ball in each bin, what's the first thing you would do? What would you do after that?

Michael Seifert
  • 9,477
4

Your answers to the first two questions are correct.

For the third question, we reduce it to a problem you know how to solve. Since each $x_k$, $1 \leq k \leq 5$, is a positive integer, the strict inequality $$x_1 + x_2 + x_3 + x_4 + x_5 < 100$$ is equivalent to the weak inequality $$x_1 + x_2 + x_3 + x_4 + x_5 \leq 99$$ If we make the substitution $y_k = x_k - 1$ for $1 \leq k \leq 5$, then each $y_k$ is a non-negative integer. Substituting $y_k + 1$ for $x_k$, $1 \leq k \leq 5$, in the weak inequality yields \begin{align*} y_1 + 1 + y_2 + 1 + y_3 + 1 + y_4 + 1 + y_5 + 1 & \leq 99\\ y_1 + y_2 + y_3 + y_4 + y_5 & \leq 94 \end{align*} which is an inequality in the non-negative integers, which you evidently know how to solve.

N. F. Taussig
  • 76,571
  • Why just subtracting $1$ is enough to make non-negative integer? I hope you mean positive integer, though not clear how. I mean that if zero value is allowed, then removing $1$ makes all values positive. I am not clear though, if negative values are possible or not. – jiten Feb 02 '21 at 05:36
  • 1
    @jiten In the third question, we were given that each $x_i > 0$. Since $x_i$ is an integer, that means $x_i \geq 1$. Hence, $y_i = x_i - 1 \geq 1 - 1 = 0$. Therefore, $y_i$ is a nonnegative integer. Negative values are not allowed. – N. F. Taussig Feb 02 '21 at 09:45