Question

A college offers 7 courses in the morning and 5 courses in the evening, find the number of ways a student can select exactly one course either in the morning or in the evening.
A. 35
B. 12
C. 40
D. 30

Answer 1

Hint: In this question it is given that a college offers 7 courses in the morning and 5 courses in the evening, so we have to find the number of ways a student can select exactly one course either in the morning or in the evening. So for this we have to know that from n number of quantity you can select r number of quantity in ${}^{n}C_{r}$ ways, where, ${}^{n}C_{r}=\dfrac{n!}{r!\cdot \left( n-r\right) !}$ and $n!=n\cdot \left( n-1\right) !$

Complete step-by-step answer:
College offers 7 courses in the morning, so a student can select one course from the 7 courses in ${}^{7}C_{1}$ ways.
Similarly, a student can select 1 course from the 5 courses in the evening in ${}^{5}C_{1}$
So by the fundamental principle of addition we can find that a student can select exactly one course either in the morning or in the evening in ${}^{7}C_{1}+\ ^{5} C_{1}$ ways, which can be written as,
${}^{7}C_{1}+\ ^{5} C_{1}$=$\dfrac{7!}{1!\cdot 6!} +\dfrac{5!}{1!\cdot 4!}$
$=\dfrac{7\cdot 6!}{1!\cdot 6!} +\dfrac{5\cdot 4!}{1!\cdot 4!}$
$=7+5$ [ since, 1! =1]
= 12 ways.
Therefore, we can say that the total number of choices is 12.
Hence the correct option is option B.

Note: In combinatorics, the rule of sum or fundamental principle of addition is a basic counting principle. Stated simply, it is the idea that if we have A ways of doing something and B ways of doing another thing and we can not do both at the same time, then there are A + B ways to choose one of the actions, and if we want to say in more simpler way then we can say whenever ‘or’ is given in the question i.e, A or B then you have to add the ways.