Final answer:
a) Mike has 84 choices of invitations to his party. b) Considering the feuding friends, Mike has 91 possibilities for invitations. c) Considering the close friends, Mike has 35 possibilities for invitations.
Step-by-step explanation:
a) In how many choices of invitations has he?
In this question, Mike wants to invite 6 friends out of 9. This is a combination problem since order doesn't matter. The number of ways to choose 6 friends out of 9 is given by the formula C(n,r) = n! / (r!(n-r)!), where n is the total number of friends and r is the number of friends to be invited. Plugging in the values, we get C(9,6) = 9! / (6!(9-6)!) = 9! / (6!3!) = 84.
b) Two of his friends are feuding and will not attend the party together. Accounting for this fact, how many possibilities has he?
In this case, we need to consider the feuding friends as separate entities. We can split the problem into two cases: when they are both invited and when only one of them is invited. If both are invited, Mike needs to choose 4 friends from the remaining 7 (since he can't invite the feuding friends). This is again a combination problem, giving us C(7,4) = 7! / (4!(7-4)!) = 7! / (4!3!) = 35 possibilities. If only one of the feuding friends is invited, Mike needs to choose 5 friends from the remaining 8. Again, this is a combination problem, providing us with C(8,5) = 8! / (5!(8-5)!) = 8! / (5!3!) = 56 possibilities. Adding up the possibilities from both cases, we have a total of 35 + 56 = 91 possibilities.
c) Two of his friends are very close and will attend the party only if invited together.
In this case, we can treat the two close friends as one entity. Now Mike needs to choose 4 friends from the remaining 7 (since he can't invite the close friends separately). This is a combination problem, giving us C(7,4) = 7! / (4!(7-4)!) = 7! / (4!3!) = 35 possibilities.