# A problem in combinatorics

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
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

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?
$8 \choose 3$ 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
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
In 95 ways because $7 \choose 1$ + $7 \choose 2$ + $7 \choose 3$ + $5 \choose 1$ + $5 \choose 2$ + $5 \choose 3$ + $3 \choose 1$ + $3 \choose 2$ + $3 \choose 3$ = $95$.