How many ways can 7 books be distributed to Robin Lily and Alexandria such that 2 are 3 Alexandria?

Here's how to solve this problem:

1. Choose the books for Alexandria:

* Alexandria gets 3 books. We need to choose 3 books out of 7. This can be done in ⁷C₃ ways, which is 7!/(3!4!) = 35 ways.

2. Distribute the remaining books:

* There are 4 books left. These need to be distributed between Robin and Lily. Since we don't care about the order they receive the books, we can think of it like this:

* Robin can get 0, 1, 2, 3, or 4 books.

* For each of these possibilities, the rest go to Lily.

3. Calculate the possibilities for Robin and Lily:

* Robin gets 0 books: There's only 1 way this can happen.

* Robin gets 1 book: There are 4 ways to choose which book Robin gets.

* Robin gets 2 books: There are ⁶C₂ = 15 ways to choose which 2 books Robin gets.

* Robin gets 3 books: There are ⁴C₃ = 4 ways to choose which 3 books Robin gets.

* Robin gets 4 books: There's only 1 way this can happen.

4. Multiply the possibilities:

* We need to multiply the number of ways to choose books for Alexandria by the number of ways to distribute the remaining books:

35 (ways to choose for Alexandria) * (1 + 4 + 15 + 4 + 1) (ways to distribute the rest) = 35 * 25 = 875

Therefore, there are 875 ways to distribute the 7 books so that Alexandria gets 3.

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