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Chapter 8: Problem 9

In how many ways can Mrs. Ford distribute ten distinct books to her ten children (one book to each child) and then collect and redistribute the books so that each child has the opportunity to peruse two different books?

Short Answer

Expert verified
The total number of ways that Mrs. Ford can distribute the 10 distinct books to her 10 children, collect them, and redistribute, ensuring each child gets to read two distinct books, is approximately \(10! \times !10 = 3628800 \times 1334961\) ways.

Step by step solution

01

Initial distribution

The initial distribution of 10 distinct books among 10 children is a simple permutation. The number of ways to do this is calculated as 10! (the factorial of 10). In general, the factorial of a number 'n' is the product of all positive integers less than or equal to 'n'. So, 10! = 10 脳 9 脳 8 脳 7 脳 6 脳 5 脳 4 脳 3 脳 2 脳 1.
02

Second round distribution

After the initial pass, Mrs. Ford collects the books and redistributes them. In the second round, it's critical that every child gets a new book they haven't read before. This is a problem of permutations without repetition. The number of ways to redistribute the books so that no child gets the same book as before is equal to the number of derangements of 10 elements (books). A derangement is a permutation of the elements of a set in which none of the elements appear in their original position. In this case, derangement of 10 elements is denoted by !10, which is approximately 1334961.
03

Total permutations

The total number of distributions (the answer to the problem) is simply the product of the two individual cases calculated in Steps 1 and 2. So, we need to multiply the number of initial distributions (10!) by the number of second round distributions (!10) to get the answer.

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