Bunuel wrote:

Robin and Terry want to invite 5 of their friends to their wedding. Robin has 7 friends, Terry has 6, and Robin and Terry have no friends in common. If at least 1 of Robin’s friends and at least 1 of Terry’s friends must be invited, how many different groups of friends could Robin and Terry invite to their wedding?

A. 462

B. 924

C. 1,260

D. 2,520

E. 151,200

We see that there are 4 cases possible: Robin invites 1, 2, 3 or 4 friends (notice that Robin can’t invite 0 friend since she has to invite at least 1 friend, and she can’t invite 5 friends since Terry has to invite at least 1 friend).

1) If Robin invites 1 friend, then Terry invites 4, and there are 7C1 x 6C4 = 7 x 15 = 105 ways to invite 5 of their friends.

2) If Robin invites 2 friends, then Terry invites 3, and there are 7C2 x 6C3 = 21 x 20 = 420 ways to invite 5 of their friends.

3) If Robin invites 3 friends, then Terry invites 2, and there are 7C3 x 6C2 = 35 x 15 = 525 ways to invite 5 of their friends.

4) If Robin invites 4 friends, then Terry invites 1, and there are 7C4 x 6C1 = 35 x 6 = 210 ways to invite 5 of their friends.

Therefore, the total number of ways to invite 5 of their friends is 105 + 420 + 525 + 210 = 1,260.

Alternate Solution:

If there were no restrictions, 5 people among a total of 6 + 7 = 13 people could have been chosen in 13C5 = 13!/(5!*8!) = (13 x 12 x 11 x 10 x 9)/(5 x 4 x 3 x 2) = 13 x 11 x 9 = 1287 ways.

From these 1287, we must subtract the sum of the number of ways in which Robin invites 5 friends and Terry invites 5 friends.

Since Robin has 7 friends, she can choose 5 friends in 7C5 = 21 ways. Similarly, since Terry has 6 friends, he can choose 5 friends in 6C5 = 6 ways. Therefore, 21 + 6 = 27 of the 1287 choices are those in which either party invites no friends. Thus, 5 friends can be invited in 1287 - 27 = 1260 ways such that either party invites at least one friend.

Answer: C

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