9514 1404 393
Answer:
a) 3003
b) 1419
Step-by-step explanation:
a) With no restrictions, the problem amounts to counting the number of combinations of 14 people taken 6 at a time. That is ...
14!/(6!(14-6)!) = 3003 . . . teams with no restrictions
__
b) With the restrictions, there are two cases:
(i) all 6 people are taken from the 12 that exclude the couple. That number of combinations is ...
12!/(6!(12-6)!) = 924
(ii) 4 people are taken from the 12 that exclude the couple, and the couple fills the last two team slots. That number is ...
12!/(4!(12-4)!) = 495
The sum of the counts of these two cases is ...
924 +495 = 1419 . . . teams with inseparable couple
