Question

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,​

Answer 1

9514 1404 393

Answer:

a) 3003

b) 1419

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