statistics

There are four children of di�erent ages in a family. Their mom purchases six di�erent gifts and decides to divide the gifts randomly between the children. Assuming that a child can receive no gift or multiple gifts, Thinking only about the number of gifts received by each child, how many di�erent possibilities are there? [for example, (3; 1; 1; 1) is one possibility where child 1 receives 3 gifts and each of the other children receive one; or (1; 3; 1; 1) is another possibility where the second child receives three gifts and the remaining children receive one; yet (0; 0; 6; 0) is another choice where the third child receives all the gifts. In this example, I have given examples of three possibilities. The problem is asking for the total number of such possibilities.]

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asked by kety
  1. This problem is the same as putting 6 items in 4 bins.
    Different ways of solving this problem would be:
    1. enumerate in a systematic way:
    0006
    0015
    0024
    0033
    ...
    6000
    and count how many you've come up with.
    Note that since the children are of different ages, and the gifts are distinct, 0006 is not the same case as, for example, 0600.

    2. the number of ways of distributing n objects into k bins can be modelled by expanding the expression (a+b+c...+k)^n. The number of terms of the expansion represents the number of ways required.
    For 6 gifts for 4 children, you would expand (a+b+c+d)^6 and count the number of terms (86) after expansion.

    (3) read up on "partitions" and "generating functions" (which is probably part of your course curriculum). You should find much information in books on combinatorics, for example your textbook. If you need recommendations, please post.

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    posted by MathMate
  2. Sorry, I had a miscount.
    The number of terms of the expansion of (a+b+c+d)^6 is 84 (not 86).

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    posted by MathMate

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