A person has 8 friends, of whom 5 will be invited to a party:
a) How many choices are there if 2 of the friends are feuding and will not attend together?
b) how many choices if 2 of the friends will only attend together?

Solution:

Given, a person has 8 friends.

5 friends are invited to a party.

Let us use the concept for combinations in solving the problem.

a) we have to find the number of choices if 2 of the friends are feuding and will not attend together

We know that \(^{n}C_{r}=\frac{n!}{r!(n-r)!}\)

Choices of selecting 5 friends out of 8 = ⁸C₅ = \(\frac{8!}{5!(8-5)!}=\frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{(5\times 4\times 3\times 2\times 1)3!}=\frac{8\times 7\times 6}{3\times 2\times 1}=56\)

Choices in which both feuding will attend = ⁶C₃=\(\frac{6!}{3!(6-3)!}=\frac{6\times 5\times 4\times 3\times 2\times 1}{(3\times 2\times 1)3!}=\frac{6\times 5\times 4}{3\times 2\times 1}=20\)

Number of choices if 2 of the friends are feuding and will not attend together = Choices of selecting 5 friends out of 8 - Choices in which both feuding will attend

= ⁸C₅ - ⁶C₃

= 56 - 20

= 36

Therefore, the number of choices is 36.

b) we have to find the number of choices if 2 of the friends will only attend together

Number of choices if 2 of the friends will only attend together = Choices of both attending the party + choices of none of them attending the party

Choices of both attending the party = ⁶C₃

= \(\frac{6!}{3!(6-3)!}=\frac{6\times 5\times 4\times 3\times 2\times 1}{(3\times 2\times 1)3!}=\frac{6\times 5\times 4}{3\times 2\times 1}=20\)

Choices of none of them attending the party = ⁶C₅

= \(\frac{6!}{5!(6-5)!}=\frac{6\times 5\times 4\times 3\times 2\times 1}{(5\times 4\times3\times 2\times 1)1!}=6\)

Now, ⁶C₃ + ⁶C₅ = 20 + 6 = 26

Therefore, the number of choices if 2 of the friends will only attend together is 26.

Therefore,

a) there are 36 choices if 2 of the friends are feuding and will not attend together.

b) there are 26 choices if 2 of the friends will only attend together.