SOLUTION: Dear math teacher, I am having difficulites solving the following math problem: In how many ways can 9 different prizes be awarded to two students so that one receives 3 an

Question 473509: Dear math teacher,
I am having difficulites solving the following math problem:
In how many ways can 9 different prizes be awarded to two students so that one receives 3 and the other 6?
So far, I got two answers but they are incorrect. The correct answer is 168.
Here is what I did so far:
First, I said to myself, this might be done this way: Permutation = 9 factorial divided by (3 factorial times 6 factorial) and that equals to 84. However, that's not the right answer. So then, I calculated: Permutation of 9 things taken 3 at a time TIMES Permution of 6 things taken 6 at time = 82.080. However, it's not the right answer either.
Would you please help me solve this problem? Thank you so much.
Yours,
I.
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
In how many ways can 9 different prizes be awarded to two students so that one receives 3 and the other 6?
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# of ways to pick a student: 2
# of ways to give 3 prizes to that student: 9C3 = 84
# of ways to give 6 prizes to the "other" student: 1
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Note: The "other" students automatically get the remaining 6 prizes.
-------
Ans: 2*84*1 = 168
=====================
Cheers,
Stan H.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Let's say there are 2 students A and B

Let's focus on how many ways to assign 3 prizes to student A.

There are 9*8*7 = 504 different ways to assign 3 prizes (out of 9) to student A where order matters. Since order does NOT matter, we divide by 3! = 6 to get 504/6 = 84. Note: this is to ignore combinations that have been counted more than once.

So there are 84 ways to assign 3 prizes to student A where order does NOT matter.

Now do the same work for student B to get 84 different ways as well. Basically repeat the above, but replace "Student A" with "Student B".

So there are 84+84=168 different ways to assign 3 prizes to one student and 6 prizes to the other.
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