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In how many ways can we distribute $7$ apples and $6$ oranges among $4$ children so that each child gets at least one apple?

I think this can be solved by using permutations because the word distribute (arrange) is given, if the word select (e.g., select a group of ..) was there I was to use combinations. Also permutations takes into consideration the order of arrangement. Am I right? Can some one explain how to solve the above problem?

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    $\begingroup$ Find the number of ways to distribute the apples, using Stars and Bars. Find the number of ways to distribute the oranges, using Stars and Bars. Multiply. $\endgroup$ – André Nicolas Feb 20 '14 at 4:39
  • $\begingroup$ @AndréNicolas stars and bars???? $\endgroup$ – techno Feb 20 '14 at 4:50
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    $\begingroup$ Please see this Wikipedia article. $\endgroup$ – André Nicolas Feb 20 '14 at 4:53
  • $\begingroup$ @AndréNicolas okay,thanks :) $\endgroup$ – techno Feb 20 '14 at 4:55
  • $\begingroup$ anyone any other answer? $\endgroup$ – techno Feb 20 '14 at 8:13
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The strategy mentioned by Andre Nicolas is also called as Balls in Urns Principle.

Suppose you have k distinguishable urns and n indistinguishable balls,there are $\dbinom{n+k-1}{k}$ ways of arranging the balls in urns.

Also,$\dbinom{n+k-1}{n}$ = $\dbinom{n+k-1}{k-1}$,which you can easily verify.

In the given question,there are 4 distinguishable children,3 indistinguishable apples and 6 indistinguishable oranges.Since every child has to have a apple,you have no choice over 4 apples.

Hence,there are $\dbinom{3+4-1}{4-1}$ ways of choosing apples and $\dbinom{6+4-1}{4-1}$ ways of choosing oranges.

Using the multiplication principle,there are $\dbinom{3+4-1}{4-1}$*$\dbinom{6+4-1}{4-1}$ of doing them together.

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Hint: Give each child their required 1 apple. Thus all you actually have to do is distribute 6 oranges and the remaining 3 apples among the 4 children. Consider each fruit separately, then multiply.

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  • $\begingroup$ okay,so let me consider the case of distributing 6 oranges to 4 children.I understand that permutations does takes into consideration the order of distribution.So cannot use i use 6P3 here right?,Please clear my doubts here $\endgroup$ – techno Feb 20 '14 at 4:47

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