Question

There are \[3\] candidates for a Classical, \[5\] for a Mathematical, and \[4\]for a Natural science scholarship. Number of ways in which these scholarships can be awarded is:
A.\[12\]
B.\[60\]
C.\[720\]
D.None of these

Answer 1

Hint: We will solve the question by using the combination formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]since we are simply told to find the number of ways and not the order of selection. Hence, we do not require the permutation formula.

Complete step by step answer:

Looking at the question, we see that there are \[3\] candidates for a Classical scholarship, \[5\] candidates for a Mathematical scholarship and \[5\] candidates for a Natural science scholarship.
Out of the \[3\] candidates for the Classical scholarship, only one of the candidates will actually be awarded the scholarship. So the number of ways in which a candidate can be awarded the Classical scholarship will be\[{}^3{C_1}\].
Out of the \[5\] candidates for the Mathematical scholarship, only one of the candidates will actually be awarded the scholarship. So the number of ways in which a candidate can be awarded the Mathematical scholarship will be\[{}^5{C_1}\].
Out of the \[4\] candidates for the Natural science scholarship, only one of the candidates will actually be awarded the scholarship. So the number of ways in which a candidate can be awarded the Natural science scholarship will be\[{}^4{C_1}\].
We are conducting one operation but we have \[3\]possible selections from the \[3\]scholarships, i.e., one from Classical, one from Mathematical and one from Natural science. So to find the total number of possibilities, we multiply.
Therefore, the total number of ways in which the scholarships can be awarded\[
   = {}^3{C_1} \times {}^5{C_1} \times {}^4{C_1} \\
   = \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{5!}}{{1!\left( {5 - 1} \right)!}} \times \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} \\
   = \dfrac{{3!}}{{2!}} \times \dfrac{{5!}}{{4!}} \times \dfrac{{4!}}{{3!}} \\
   = \dfrac{{3 \times 2!}}{{2!}} \times \dfrac{{5 \times 4!}}{{4!}} \times \dfrac{{4 \times 3!}}{{3!}} \\
   = 3 \times 5 \times 4 \\
   = 60 \\
\]
Thus, the answer is option B.
Note: This question can also be solved directly. As Classical scholarship can be awarded to any of the \[3\] candidates, Mathematical scholarship can be awarded to any of the \[5\]candidates and Natural science scholarship can be awarded to any of the \[4\]candidates.
Therefore, the number of ways in which the scholarships can be awarded\[ = 3 \times 5 \times 4 = 60\]ways.