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

Hint:
Here the possibility or ways of getting the gold medal is equal to the number of students awarded. Use permutations or combinations for finding the different ways of distribution or selection.

Complete step by step solution:
Here it is given that,
1) 4 students for physics
2) 6 students for chemistry
3) 7 students for mathematics
are for gold medals.
We need to find the ways in which these students will get gold medals.
ways in which these gold medals be awarded \[ = {}^{4}{C_1} + {}^{6}{C_1} + {}^{7}{C_1}\]
\[
   \Rightarrow \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} + \dfrac{{6!}}{{1!\left( {6 - 1} \right)!}} + \dfrac{{7!}}{{1!\left( {7 - 1} \right)!}} \\
   \Rightarrow \dfrac{{4!}}{{3!}} + \dfrac{{6!}}{{5!}} + \dfrac{{7!}}{{6!}} \\
   \Rightarrow 4 + 6 + 7 \\
   \Rightarrow 17 \\
\]
So there are 17 ways in which the medal can be awarded.

So option A is correct.

Additional information:
1) The combination is used when we need to do selection especially.
2) Like from similar things or dissimilar things, we have to select in different ways, patterns.
3) It is used in conditions of picking balls, picking a card, selecting a committee, forming a team.

Note:
Here every student from the mentioned stream will get the medal. So we have a number of students is equal to that of ways of medal distribution.