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There are 6 indistinguishable oranges, and 4 unique apples. There are 7 people. How many ways can each person get 1 piece of fruit? Then, say Person A requires an apple, how many ways can this be done now?

This was my intuition:

The first part must be considered in 4 cases (how many apples are distributed). We know there must be at least 1 apple given out as there are only 6 oranges.

Case: 1 apple given out

  • Select who gets apple (7 ways)
  • Select which apple (4 ways)
  • Give everyone else oranges (1 way)

Case: 2 apples given out

  • Select who gets first apple (7 ways)
  • Select which apple (4 ways)
  • Select who gets second apple (6 ways)
  • Select which apple (3 ways remaining)
  • Give everyone else oranges (1 way)

Case: 3 apples given out

  • Select who gets first apple (7 ways)
  • Select which apple (4 ways)
  • Select who gets second apple (6 ways)
  • Select which apple (3 ways remaining)
  • Select who gets third apple (5 ways)
  • Select which apple (2 ways)
  • Give everyone else oranges (1 way)

Case: 4 apples given out

  • Select who gets first apple (7 ways)
  • Select which apple (4 ways)
  • Select who gets second apple (6 ways)
  • Select which apple (3 ways remaining)
  • Select who gets third apple (5 ways)
  • Select which apple (2 ways)
  • Select who gets fourth apple (4 ways)
  • Give everyone else oranges (1 way)

Then by addition, add them up for the total number of ways.

Now, I'm starting to think that this might be no good because of the dependence of apple selection. Instead, while it still may be in cases, for multiple apples (say 3) you would say

  • Choose the 3 people to get the apples ($7\choose3$ ways)
  • Choose which 3 apples you want ($4\choose3$ ways)
  • Give out oranges to rest (1 way)
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  • $\begingroup$ and then by continuing the logic from above, the answer to the second part would be to give an apple to person A (4 ways) and then treat the question as having 3 apples and 6 oranges with 6 people. $\endgroup$ – Neurax Feb 27 '14 at 5:30
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Let us count the number of ways to hand out say $3$ apples (the others can be done the same way).

As in your second suggestion, the apples can be chosen in $\binom{4}{3}$ ways, and the apple-getters can be chosen in $\binom{7}{3}$ ways. Now line up the apple-getters in order of height, or Student Number. We have $3$ choices for which apple goes to the shortest student, and then $2$ choices for the next, then $1$ "choice," for a total of $\binom{4}{3}\binom{7}{3}3!$.

Alternately, choose the apple-getters, and line them up in order of height. There are $4$ ways to decide which apple the shortest person gets, and so on down to $2$, for a total of $\binom{7}{3}(4)(3)(2)$.

Alternately, we can choose the apples, and line them up in order of redness. That gives answer $\binom{4}{3}(7)(6)(5)$.

Remark: Note that your first method overcounted: there is no such thing as a first person, or apple.

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  • $\begingroup$ Yeah.. Thank you again @AndréNicolas. I wish I had this insight during the exam, but now I know for the final. $\endgroup$ – Neurax Feb 27 '14 at 5:55
  • $\begingroup$ You are welcome. Probably the only way one really learns about overcounting is to have done it. $\endgroup$ – André Nicolas Feb 27 '14 at 5:57

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