SOLUTION: A man wishes to give his 12 books to his 3 children, so that the first shall receive 5, the second 4, and the third 3. In how many ways can he give the books?

Question 1157327: A man wishes to give his 12 books to his 3 children, so that the first shall receive 5, the second 4, and the third 3. In how many ways can he give the books?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13216)   (Show Source): You can put this solution on YOUR website!

The rules are not clear; so I will assume the easy solution.

The number of ways he can give the books is the number of ways to arrange the numbers 3, 4, and 5 in a row: 3!=6 ways.
1st 2nd 3rd child child child ------------------- 3 4 5 3 5 4 4 3 5 4 5 3 5 3 4 5 4 3

Answer by ikleyn(52915)   (Show Source): You can put this solution on YOUR website!
.

He can select 5 books for the first child by ways = = 792 ways. (the order is not important). He then can select 4 books from remaining 12-5 = 7 books for the second child by = ways = = 35 ways. The rest of the books go to the third child, with no choice. In all, there are 792*35 = 27720 different ways. ANSWER

Solved.

------------------

On Combinations, see introductory lessons
    - Introduction to Combinations
    - PROOF of the formula on the number of Combinations
    - Problems on Combinations
    - OVERVIEW of lessons on Permutations and Combinations
in this site.

Also, you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Combinatorics: Combinations and permutations".

Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.

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