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8 subjects need to be given to 4 students. In how many ways can it be done so that the third student gets an odd number of subjects.

I tried combination with repetition $9 \choose 7$+$7 \choose 5$+$5 \choose 3$+$3 \choose 1$

but i'm not quite sure this is the right way to solve the problem. Can anyone help me? Thanks

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    Does everyone get at least one subject? Are subjects like presents, or could Alice and Bob both get Geometry among their subjects?2012-06-21

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Hint: If you start with the third student, there are $\binom 81$ ways to give one subject, then $3^7$ ways to give the remaining $7$ to the other two, as each of the other $7$ subjects can be given to any one. Can you extend this to other numbers of subjects for the third student?

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    @10001a: Exactly. You multiply them because for each class you give the third the 3^7 ways to give the rest of the classes are all the possibilities. You add the different numbers of classes for the first because those sets are disjoint: having decided which classes to give the first, you have a distinct set to give the rest.2012-06-21
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In 95 ways because $7 \choose 1$ + $7 \choose 2$ + $7 \choose 3$ + $5 \choose 1$ + $5 \choose 2$ + $5 \choose 3$ + $3 \choose 1$ + $3 \choose 2$ + $3 \choose 3$ = $95$.

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    I imagine that couple of lines of explanation would be useful to anyone coming across this...2014-03-31