Question

There are 4 students for Physics, 6 students for Chemistry and 7 students for Mathematics gold medals. In how many ways can one of these gold medals be awarded?
A) 17
B) 11
C) 7
D) None of these

Answer 1

Here we have that the possibility or ways of getting the gold medal is the same as the number of students awarded. So, we will use the formula of permutations, that is, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, where $n$ is the total number of object and $r$ is the number required to find the required value.

Complete step by step solution:
We are given that there are 4 students for Physics, 6 students for Chemistry and 7 students for Mathematics gold medals.
So, we have to find the ways in which these above students will get gold medals.
We know that the formula of combinations, that is, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, where $n$ is the total number of object and $r$ is the number required.
For the subject Physics we will have $n = 4$.
Since only one will receive the gold medal, so we have $r = 1$.
Computing the value of $n$ and $r$ in the above formula of combination to find the ways one of the gold medal be awarded for Physics, we get

$$\Rightarrow \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} \\\ \Rightarrow \dfrac{{4!}}{{1! \cdot 3!}} \\\$$

Simplifying the factorial of the above equation, we get

$$\Rightarrow \dfrac{{4 \times 3!}}{{1 \cdot 3!}} \\\ \Rightarrow 4{\text{ .......eq.(1)}} \\\$$

For the subject Chemistry we will have $n = 6$.
Since only one will receive the gold medal, so we have $r = 1$.
Computing the value of $n$ and $r$ in the above formula of combination to find the ways one of the gold medal be awarded for Chemistry, we get

$$\Rightarrow \dfrac{{6!}}{{1!\left( {6 - 1} \right)!}} \\\ \Rightarrow \dfrac{{6!}}{{1! \cdot 5!}} \\\$$

Simplifying the factorial of the above equation, we get

$$\Rightarrow \dfrac{{6 \times 5!}}{{1 \cdot 5!}} \\\ \Rightarrow 6{\text{ .......eq.(2)}} \\\$$

For the subject Mathematics we will have $n = 4$.
Since only one will receive the gold medal, so we have $r = 1$.
Computing the value of $n$ and $r$ in the above formula of combination to find the ways one of the gold medal be awarded for Mathematics, we get

$$\Rightarrow \dfrac{{7!}}{{1!\left( {7 - 1} \right)!}} \\\ \Rightarrow \dfrac{{7!}}{{1! \cdot 6!}} \\\$$

Simplifying the factorial of the above equation, we get

$$\Rightarrow \dfrac{{7 \times 6!}}{{1 \cdot 6!}} \\\ \Rightarrow 7{\text{ .......eq.(3)}} \\\$$

Adding the equation (1), (2) and (3) to find the ways in which the medal can be awarded, we get

$$\Rightarrow 4 + 6 + 7 \\\ \Rightarrow 17 \\\$$

Therefore, we have 17 ways in which they can be awarded.
Hence, option A is correct.

Note:
Some students mistakenly use the formula of permutations, ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ instead of the formula of permutations, so one should kept in mind such formulas while solving problems like this. We have to know that every student from the mentioned stream will get the medal. While counting the possible combination factors in the above question, the factors could be chosen such that the factor on the left is always greater than the factor on the right, never less. This will help in rectifying the redundant cases.