A team of 6 people is to be chosen from 8 men and 6 women. Find the number of different teams that may be chosen if (a) there are no restrictions,
(b) there are a husband and wife who must not be seperated,

9514 1404 393

  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

for future students the answer is b

hope i

-lavira: )