Question: The number of ways that 12 prizes can be divided among 4 students so that each may have 3 prizes is:...

Question

The number of ways that 12 prizes can be divided among 4 students so that each may have 3 prizes is:
A.15,400
B.15,300
C.15,151
D.369,600

Answer 1

Solution

We will use the method of combinations to calculate the number of ways such that 12 prizes can be divided among 4 students so that each may have 3 prizes and in the last, we will multiply all the possible conditions to calculate the final number of ways because all of the ways altogether forms the total ways not individually.

Complete step-by-step answer:
We are given that 12 prizes need to be distributed among 4 students so that each can have 3 prizes.
Therefore, each will get 3 prizes and there will be no remaining prize. Then, 12 prizes can be distributed among 4 students as:
The 1st student will get 3 prizes from a total of 12 prizes. Therefore, total ways in which this can be distributed are ${}^{12}{C_3}$ .
Now, there are 9 remaining prizes from a total of 12 prizes for the remaining 3 students.
For 2nd students, he will receive 3 prizes from a total of 9 prizes. The total ways in which it can be done are ${}^9{C_3}$ .
Now, there are 6 remaining prizes from a total of 12 prizes for the remaining 2 students.
3rd students will receive 3 prizes from the remaining 6 prizes. The total number of ways to do so are ${}^6{C_3}$
There will be only 3 prizes left and they will be given to 4th students in ${}^3{C_3}$ ways.
We know that all these prizes will be distributed to the students simultaneously, hence, they will be multiplied with each other to get the total number of ways.
$\Rightarrow$ total number of ways = ${}^{12}{C_3} \times {}^9{C_3} \times {}^6{C_3} \times {}^3{C_3}$
$\Rightarrow {}^{12}{C_3} \times {}^9{C_3} \times {}^6{C_3} \times {}^3{C_3} = 220 \times 84 \times 20 \times 1 = 369,600$
Therefore, the total number of ways to distribute 12 prizes among 4 students equally is 369,600.
Hence, option(D) is correct.

Note: In such problems, students might get confused between what to use i.e., either permutations or combinations. Also, you need to take care of multiplication or addition to get the final answer.