Question

A man has 10 friends and wants to arrange them on chairs taking 6 friends at a time. Find the number of arrangements such that:
a. Two particular friends are always taken.
b. Four particular friends are never taken.

Answer 1

Hint: For solving this question we should know about the permutation and combination concept. In this problem we will use the permutation for finding the number of ways in which the friends can be arranged. And after getting the answer for the first condition, take the number of friends as 8 in the next because only two were taken in the first condition out of 10.

Complete step by step answer:
According to the problem, it is asked to find the number or arrangements and two situations are given to us for finding the number of arrangements.
As we know, permutations are ordered combinations. It means where the order does not matter, there it is a permutation. And permutations have also two types, one is repeated or repetition is allowed and the second is no repetition. The permutations with repetition are always the easiest to calculate. And the permutation with no repetition reduces the number of available choices each time. So, if we see our question, then:
Number of friends a man has is 10. 6 friends are taken at a time in which two particular friends are always taken, so these two friends are arranged in ${}^{6}{{P}_{2}}$ ways.
Now the remaining 4 friends can be arranged by using remaining (10 - 2) friends in P (10 -2, 6 - 2) ways = ${}^{8}{{P}_{4}}$.
Total number of ways = ${}^{8}{{P}_{4}}\times {}^{6}{{P}_{2}}=50400$.
So, the total number of ways is 50400.

Note: While solving this type of questions you have to ensure that we will use the permutation or combination here and, in the permutation, it is necessary to find out if there are any events that are repeating or not repeating.