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Chapter 1: Problem 12

In how many ways can 12 different books be distributed among four children so that (a) each child gets three books? (b) the two oldest children get four books each and the two youngest get two books each?

Short Answer

Expert verified
For part (a), there are \(4 * {12 \choose 3} * 3!\) ways, and for part (b), there are \({12 \choose 4} * 4! * {8 \choose 4} * 4! * {4 \choose 2} * 2! * {2 \choose 2} * 2!\) ways.

Step by step solution

01

Part (a)

We first need to choose 3 books out of 12, which can be done in \({12 \choose 3}\) ways. Next, these books can be arranged in 3! ways. Furthermore, we have 4 children to choose from. Therefore, the total number of ways is \(4 * {12 \choose 3} * 3!\).
02

Part (b)

In this case, we first choose 4 books separately for two oldest children, which can be done in \({12 \choose 4}\) ways each, and arranged in 4! ways each. Similar operation is done for two youngest children but choosing 2 books each. So, the total number of ways is \({12 \choose 4} * 4! * {8 \choose 4} * 4! * {4 \choose 2} * 2! * {2 \choose 2} * 2!\).

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Most popular questions from this chapter

a) Find the number of ways to write 17 as a sum of \(1^{\prime} s\) and 2 's if order is relevant. b) Answer part (a) for 18 in place of 17 . c) Generalize the results in parts (a) and (b) for \(n\) odd and for \(n\) even.

How many ways are there to place 12 marbles of the same size in five distinct jars if (a) the marbles are all black? (b) each marble is a different color?

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