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Consider the following equation $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 20$$

a) Count the number of integer solutions of the equation under the condition that $x_i \ge 0$ for $i=1, 2,\ldots 6$.

I found the answer, it is $53130$.

b) Suppose that we choose one the solutions from part (a) at random. What is the probability that this solution will satisfy the condition that $0 \le x_i \le 8$ for $i = 1, 2,\ldots 6$. (Hint: Define $A_i$ the event that $x_i > 8$)

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closed as off-topic by Matthew Conroy, ronno, Jonas Meyer, J. W. Perry, Claude Leibovici May 19 '15 at 4:55

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  • $\begingroup$ Can you expand on what you've done so far, or what you're having problem with? $\endgroup$ – Demosthene May 18 '15 at 20:45
  • $\begingroup$ for part a) I did ( n + r - 1)C(r - 1). I got 25C5. For part b) i don't know where to start. If I let Ai be the event that xi > 8, I don't understand how i would find the probability for the condition given. $\endgroup$ – user234942 May 18 '15 at 20:47
  • $\begingroup$ $A_i$ is the complement of the probability you're trying to derive. $\endgroup$ – Demosthene May 18 '15 at 20:51
  • $\begingroup$ Ok, but i don't seem to understand how I would go about this problem. Am i still working with combinations here? $\endgroup$ – user234942 May 18 '15 at 20:54
  • $\begingroup$ $P(0\le x_i\le 8)=1-P(x_i>8)=1-A_i$. And you can find $A_i$ in a similar fashion as in question a). $\endgroup$ – Demosthene May 18 '15 at 20:55
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Consider we have an arrangement of 20 balls and 5 dividers - representing six variables to contain 20 integer solutions.

How many ways are there to place the dividers between the balls, when more than one divider can be placed between any two balls ? This is the permutation of 25 items composed of a group of 20 and a group of 5 indistinguishable items.

$$\lvert \Omega\rvert = {^{25}\mathsf C_{5}}$$


For part (b), the favoured event is: $A= \bigcap_{i=1}^6 \{0\leq x_i\leq 8\}$

Then its complement is: $A^\complement = \bigcup_{i=1}^6\{x>8\}$

That is, the complement is the event that at least one partition at least 9 balls.   To count this we first remove 9 balls and count the ways to arrange partitions between the rest.   Then multiply the result by the count of ways to select one of six numbers to add back the removed block of balls.   However we have over counted so must use the Principle of Inclusion Exclusion to account for cases where two partitions each have at least 9 .

$$\lvert A^\complement\rvert = {^6\mathsf C_1}\cdot{^{(20+5-9)}\mathsf C_5} -{^6\mathsf C_2}\cdot{^{(20+5-18)}\mathsf C_5}$$

Can you continue from here?

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  • $\begingroup$ If i remove 8 from 20, then the ways to arrange the partitions is 17C5. Is this correct? Then i multiply this value by (8+5)C5 ? After I divide this product by 25C5 I don't get a decimal value. Am i missing something? $\endgroup$ – user234942 May 19 '15 at 0:48
  • $\begingroup$ No, @user234942 , you choose which of the 6 variables to add back unto. That is $^6\mathsf C_k$. $\endgroup$ – Graham Kemp May 19 '15 at 0:58
  • $\begingroup$ Would it be 6C5? $\endgroup$ – user234942 May 19 '15 at 1:05
  • $\begingroup$ Would i continue up to 6C6? $\endgroup$ – user234942 May 19 '15 at 1:28
  • $\begingroup$ @user234942 You have only 20 balls; you can not reserve more than 2 lots of nine. $\endgroup$ – Graham Kemp May 19 '15 at 1:37

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