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how many ways are there to distribute seven balls into five boxes, where each box must have at least one ball in it, if (a) both the balls and boxes are unlabeled? (b) the balls are labeled, but the boxes are unlabeled? (c) both the balls and boxes are labeled?

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Final answer:

There are 792 ways to distribute seven unlabeled balls into five unlabeled boxes, 1 way to distribute seven labeled balls into five unlabeled boxes, and 78125 ways to distribute seven labeled balls into five labeled boxes.

Step-by-step explanation:

(a) If both the balls and boxes are unlabeled, we can use the concept of stars and bars to solve this problem. Let's imagine that we have 7 stars representing the balls and 5 bars representing the boxes. We need to place the 5 bars in between the 7 stars in such a way that each box has at least one star. Therefore, we have to find the number of ways to arrange the 7 stars and 5 bars, which is equivalent to finding the number of ways to choose 5 positions for the bars from a total of 12 positions (7 stars + 5 bars). This can be calculated using the combination formula as C(12, 5) = 792 ways.

(b) If the balls are labeled but the boxes are unlabeled, we can use the concept of distributing labeled balls into unlabeled boxes. In this case, each ball can be placed into any of the 5 boxes without any restrictions. Therefore, the number of ways to distribute the balls is simply equal to the number of ways to choose 5 boxes out of 5, which is 1.

(c) If both the balls and boxes are labeled, we can think of this problem as distributing labeled balls into labeled boxes. Each ball has 5 choices of boxes to be placed in, and since there are 7 balls, the number of ways to distribute the balls is 5^7 = 78125.

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