Question

We wish to select 6 persons from 8, but if person A is chosen, then B must be chosen. In how many ways can selection be made?
(A) 20
(B) 22
(C) 24
(D) 26

Answer 1

Hint: We will use the combination's formula to find the number of ways the selection can be made. We can have an instance when person A is not chosen, so the number of ways of selection is $${}^6{C}_6$$. There can be another instance when person A is chosen that follows that B is also chosen, so the number of ways of selection is $${}^6{C}_4$$. There can be one more possibility that B is chosen but person A is not chosen, then the number of ways of selection is $${}^6{C}_5$$. We will add up all the possibilities to find the total number of ways of selection.

Complete step by step solution:
According to the given question, we have to find the number of ways the selection is possible with a condition given.
Here we can see that the order is not mandatory during the selection, so in such cases when the order is not compulsory we will use a combination formula.
We have,
$${}^n{C}_r={\displaystyle \frac{n!}{r!(n-r)!}}$$
Where $${}^n{C}_r$$ is the number of combination,
$$n$$ is the total number of objects in a given set,
$$r$$ is the number of objects required from the same set.

So, we have to choose 6 persons with the condition being that if person A is chosen then person B should also be chosen. Based on this given condition, we can have some possibilities, we will take them as different cases.
Case 1: when both persons A and B are not chosen,
Then the number of ways of selection is $${}^6{C}_6={\displaystyle \frac{6!}{6!(6-6)!}}$$
$${\Rightarrow }^6{C}_6={\displaystyle \frac{6!}{6!0!}}=1$$
Case 2: when neither of the persons A and B are chosen,
Then the number of ways of selection is $${}^6{C}_4={\displaystyle \frac{6!}{4!(6-4)!}}$$
Case 3: when only person B is chosen,
Then the number of ways of selection is $${}^6{C}_5={\displaystyle \frac{6!}{5!(6-5)!}}$$
$${\Rightarrow }^6{C}_5={\displaystyle \frac{6!}{5!1!}}={\displaystyle \frac{6\times 5!}{5!}}=6$$
So, the total number of ways $${=}^6{C}_6{+}^6{C}_4{+}^6{C}_5$$
$$=1+15+6=22$$ `
Therefore, the number of ways of selection is (B) 22.


So, the correct answer is “Option B”.

Note: The combination formula applied should carefully and in separate steps to avoid mistakes. Since the cases made were for the same condition, so we added the values of the above mentioned three cases. And because the order was not the criteria, so we used combinations, if particular order were to be followed, then we would have used permutation.