Number of ways 9 prizes can be given to 5 students if one particular student receives 4 prizes and the rest of the students can get any number of prizes is?
My Attempt:
I got the possibilities for distribution of prizes as,
$(4,0,1,1,3)$
$(4,1,1,1,2)$
$(4,0,0,3,2)$
Solving further includes permutations of the prizes,
$5![\frac{9!}{4!2!3!} + \frac{9!}{4!3!2!} + \frac{9!}{4!2!3!2!}]$
However, solving this gives me a wrong answer.
Any help would be appreciated.
On
Ok so there are 9 prizes 5 students but one student got 4 prizes [Assume he is particular one,Also assume all prizes are same] There are 5 Prizes and 4 Students. Only one student get 5 prizes(5,0,0,0)+(0,5,0,0)+(0,0,5,0)+(0,0,0,5) So number of ways=4 Now one student got 4 and one got 1 rest got nothing =factorial_r(4)/factorial_r(2) That is 12 ways Now one student got 3,second got 1,third one got 1 Now it is factorial_r(4)/factorial_r(1) That is 24 ways Now 3 all got 1 prize and one got 2 it is factorial_r(4) That is 24 ways Total= 4+12+24+24 That is 64 ways. If 4 prizes won by student is not particular then 4*64= 256 ways If all prizes were different then it is different story!
Following your approach, the distribution of prizes (we assume that they are distinguishable) should be $$({\bf 4},1,1,1,2),({\bf 4},0,1,1,3),({\bf 4},0,1,2,2),({\bf 4},0,0,1,4),({\bf 4},0,0,2,3),({\bf 4},0,0,0,5)$$ and including the permutations of the remaining $5$ prizes and of the other $4$ students we obtain $$\binom{9}{4}\left(\frac{4!}{3!1!}\cdot \frac{5!}{1!1!1!2!}+ \frac{4!}{1!2!1!}\cdot \frac{5!}{0!1!1!3!}+\dots+\frac{4!}{3!1!}\cdot \frac{5!}{0!0!0!5!}\right)=129024.$$ The same result can be obtained in a simpler way as $$\binom{9}{4}\cdot 4^5=129024$$
where $\binom{9}{4}$ is the number of ways to assign $4$ prizes to the particular student and $4^5$ is the number of ways to give each of the remaining $5$ prizes to one of the other $4$ students.