Question

A person has 12 friends and he wants to invite 8 of them to a birthday party. Find how many times 3 particular friends will always attend the parties.

Answer 1

Hint- To solve this question, we need to know the basic theory related to the permutation and combination. So, before solving this question you need to first recall the basic understanding of this chapter. For example, if we need to select 8 friends out of 12 friends. In this case, this can be done in${}^{12}{{\text{C}}_8}$.
Formula used- ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$

Complete step by step solution:
As given in the question,
Total no. of friends = 12
As we know,
for selecting r things out of n-
${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
Now, as per our requirement in our question.
Total 12 friends are there and 8 has to be invited and 3 particular has to be invited every time
So, selecting three particular friends = ${}^3{{\text{C}}_3}$
So, he has to choose to remain 5 in remaining 9 friends
and that is possible in ${}^9{{\text{C}}_5}$= $\dfrac{{{\text{9!}}}}{{{\text{[5}}!{\text{(9 - 5)}}!{\text{]}}}}$
= $\dfrac{{{\text{9!}}}}{{{\text{[5}}!{\text{ 4}}!{\text{]}}}}$
= $\dfrac{{{\text{5!}} \times 6 \times 7 \times 8 \times 9}}{{{\text{[5}}!{\text{ 4}}!{\text{]}}}}$
= 126
So, for attending 3 particular friends always (in the parties) = ${}^3{{\text{C}}_3}$ $\times$126
= 126 [${}^3{{\text{C}}_3}$= 1]
Therefore, 126 times 3 particular friends will always attend the parties in this case.

Note- We need to remember this formula for selecting r things out of n.
${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
where n! means the factorial of n.
factorial is always defined for a positive integer n, denoted by n! and it is the product of all consecutive positive integers less than or equal to n.
for example, $5!{\text{ = 5}} \times {\text{4}} \times {\text{3}} \times {\text{2}} \times 1$