Question

How many ways are there to distribute five balls into seven boxes if each box must have at most one ball in it if. the balls are labeled, but the boxes are unlabeled?

Answer 1

There are 21 ways to distribute five labeled balls into seven unlabeled boxes, with at most one ball per box.

Step-by-step explanation:

The question is asking how many ways there are to distribute five labeled balls into seven unlabeled boxes where each box can have at most one ball. This is a combinatorial problem in which we are looking to count arrangements. Since the boxes are unlabeled, the order in which we place the balls into boxes does not matter; however, the balls are distinct from one another due to labels.

To solve this problem, we recognize that it is a case of choosing which five boxes (out of the seven available) will receive a ball.

The number of ways to choose the five boxes is the combination of seven taken five at a time, which is calculated using the binomial coefficient formula C(n, k) = n! / (k! (n-k)!), where n is the total number of items, and k is the number of items to choose.

Therefore, the number of ways to place the five labeled balls into seven unlabeled boxes is C(7, 5) which is equal to 7! / (5! * (7-5)!) = 7*6 / (2*1) = 21 ways.