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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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1. Subjects are distinguishable from each other, therefore you need to used ordered counts instead of unordered counts. 2. Once you assign subjects to the third student, you need to also figure out, in how many ways can the remaining subjects be distributed amongst the other three. –  TenaliRaman Jun 21 '12 at 13:43
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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? –  André Nicolas Jun 21 '12 at 14:33
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2 Answers

up vote 2 down vote accepted

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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$ 8 \choose 3$ ways to give 3 subjects to the third student and 3^5 ways to give the remaining 5 to the other three students etc. Note: There are 4 students not 3 –  10001a Jun 21 '12 at 13:53
    
@10001a: you are right. I have fixed it. Now you just add the four possibilities together and you are there. –  Ross Millikan Jun 21 '12 at 13:59
    
One more question though i multiply $8 \choose 1$ with 3^7? ( $8 \choose 3$ with 3^5 and so on) Thanks a lot –  10001a Jun 21 '12 at 14:05
    
@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. –  Ross Millikan Jun 21 '12 at 14:28
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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... –  Joe Tait Mar 31 at 14:27
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