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"How many solutions are there for x+y+z+w+f=15 if all x, y, z, w, f are all integers greater than or equal to one, i.e., x,y,z,w,f>=1."

Is this a question that will be using permutations or combinations? Any help would be greatly appreciated!

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Consider $15$ balls in a row. We decide to put $4$ sticks, each between two balls. We see that for each way putting $4$ sticks, we find a solution for $(x,y,z,w,f)$. Thus, there are $\binom{14}{4}$ ways to put $4$ sticks between balls, or $\binom{14}{4}$ integer solutions for $(x,y,z,w,f)$ satisfying the condition.

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Consider each of $x, y, z, w, f$ to be a box. You can interpret this as adding 15 balls to five boxes such that each box has one ball. Because each box must have at least one ball (the constraint $x, y, z, w, f \ge 1$), we can add one ball to each bin. We now need to add the remaining 10 balls to the 5 bins. Using a stars-and-bars argument, we need 4 divisors to separate the boxes. We have 14 objects (10 balls, 4 dividers), so there are ${14 \choose 4}$ ways to select $(x,y,z,w,f)$.

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