Question

A person wishes to make up as many different parties as he can out 20 friends, each party consisting of same number. How many he should invite at a time? In how many ways the same man would be found?

Answer 1

Hint:The above-given question is the question of permutation and combination. So, to find the number of friends he should invite to each party so that he can make maximum parties is given by the maximum value ${}^{n}{{C}_{a}}$and here n=10. So, we have to find r for which ${}^{n}{{C}_{a}}$is maximum. It is maximum when $a=\dfrac{n}{2}$, if n is even. To solve the second part, we will say that friend A is fixed so we have to choose the remaining (a-1) friend for the party out of (n-1) (i.e.20-1) friends.

Complete step by step answer:

We know that the above-given question is a question of permutation and combination.
Also, we can see that in the question it is given that the person has 20 friends and he wants to make as many parties as he can by taking the same number of friends.
Since he is inviting the same number of friends for each party so, let us assume that he invites ‘a’ friend for each party and ‘a’ must be less that or equal to 20 because he has maximum 20 friends. So, a total number of parties he can make by inviting ‘a’ friend each time will be equal to ${}^{20}{{C}_{a}}$ because he has to select ‘a’ friend each time out of 20 and he can invite the same friend each time for party. And, we know that this can only be done by selecting all possible combinations we can have.
Now, we know that person wants to make as many parties as he can by inviting the same number of friends each time. So, we have to maximize${}^{20}{{C}_{a}}$and we know that ${}^{n}{{C}_{a}}$is maximum when ‘a’ is its middle term of ${}^{n}{{C}_{a}}$.
And, we also know that we will get maximum term when $a=\dfrac{n}{2}$ if n is even.
And, we can see that in${}^{20}{{C}_{a}}$, n=20 and since 20 is even, so we will get the maximum term when$a=\dfrac{20}{2}$
$\Rightarrow a=10$
Hence, the person should invite 10 friends each time for the party.
Now, in the second part of the question, we have to find that how many times the person has invited the same friend (i.e. we have to find in how many parties the same friend will be found.)
Since the party is possible only if he invites at least 1 friend each time and the person have to invite exactly 10 friends each time for the maximum number of parties.
Let us assume that the person’s friend ‘A’ is fixed and is found in the maximum number of parties.
So, now the person has to select only the remaining 9 friends in which the friend ‘A’ is actually present.
Hence, we have to find the number of ways in which the remaining 9 friends can be selected out of the remaining 19 friends as we have fixed 1 friend whose name is ‘A’.
So, we can say that number of ways in which the remaining 9 friends can be selected is equal to ${}^{19}{{C}_{9}}$.
Hence, we can say that the same man would be found in the ${}^{19}{{C}_{9}}$ parties.

Note:
 Students are required to note that the maximum term of ${}^{n}{{C}_{a}}$ is the middle term. So, while finding the middle term of the ${}^{n}{{C}_{a}}$ students are required to check whether n in ${}^{n}{{C}_{a}}$ is even or odd. If n is even then there is 1 middle term and it is given by $a=\dfrac{n}{2}$and if n is odd then there is 2 middle term $a=\dfrac{n-1}{2}$ and $a=\dfrac{n+1}{2}$, so to find maximum term of ${}^{n}{{C}_{a}}$we have to explicitly check whether we get maximum at $a=\dfrac{n-1}{2}$ or at $a=\dfrac{n+1}{2}$.