Question

From a group of 12 students, 8 are to be chosen for an excursion. There are 3 students who decide that either of them will join or none of them will join. In how many ways can the 8 be chosen?

Answer 1

Hint: In this question, it is given that there are a total 12 students out of which 8 are to be chosen for an excursion. In the group of 12, 3 are such students which decide that either all of them will join the recursion or none will join. So, we need to find all the possible ways in which 8 students are to be chosen by using combination.

Complete step by step solution:
Let’s discuss the problem now.
As we know that combination is a method which determines the number of possible arrangements in a collection of items where the order of the selection does not matter. The formula for combination is:
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Where ${}^{n}{{C}_{r}}$ = number of combinations, n = total number of objects in the set and r = number of choosing objects from the set.
As per question, there are 12 students in a group out of which 8 are to be chosen for an excursion. Out of 12, 3 are such students which decided that either all of them will join the excursion or none of them will join. We have to find all the possible ways in which 8 students need to be chosen for an excursion. So there are 2 cases in which 8 students have to be chosen.
First case: If all three students will join:
If all the three of them decided to join, then the remaining will be 12 – 3 = 9 from the group of 12 students and if 8 are to be chosen then 3 will be fixed, so now remaining 8 – 3 = 5 students has to be chosen from 9 students. So the combination will be:
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Put the values of n = 9 and r = 5:
$\Rightarrow {}^{9}{{C}_{5}}=\dfrac{9!}{5!\left( 9-5 \right)!}=\dfrac{9!}{5!\left( 4 \right)!}$
5! Will be cancelled with the numerator, we will get:
$\Rightarrow {}^{9}{{C}_{5}}=\dfrac{9\times 8\times 7\times 6}{4\times 3\times 2}$
Now, multiply all the terms:
 $\Rightarrow {}^{9}{{C}_{5}}=\dfrac{3024}{24}$
Reduce the fraction into simplest form:
$\Rightarrow {}^{9}{{C}_{5}}=126$
Second case: If none of the students will join:
If none of the students will join then those three students will be removed from a group of 12 students i.e. 12 – 3 = 9. Now from 9 students 8 will be chosen for the excursion. So the combination will be:
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Put the values of n = 9 and r = 8:
$\Rightarrow {}^{9}{{C}_{8}}=\dfrac{9!}{8!\left( 9-8 \right)!}=\dfrac{9!}{8!\left( 1 \right)!}$
So,
$\Rightarrow {}^{9}{{C}_{8}}=9$
The total number of ways in which 8 students can be chosen are:
$\Rightarrow {}^{9}{{C}_{5}}$+ ${}^{9}{{C}_{8}}$
$\Rightarrow $126 + 9 = 135 ways

Note: Factorial of a number is represented by n!. For example we have to find the factorial of 5, it will be 5! = $5\times 4\times 3\times 2\times 1$ which is equal to 120. In the same way we have obtained in the answer above. Remember the formula for combinations otherwise it won’t be possible to find the total number of ways in which 8 students can join the excursion.