jimmy.student wrote:A committee of three people is to be chosen from four married couples. What is the number of differenct committees that can be chose if two people who are married to each other cannot both serve on the commitee?

16

24

26

30

32

Hi Jimmy.

People are (say) A1, A2, B1, B2, C1, C2 and D1, D2 (same letter = couple) .

(1) Choose 3 "letters" (couples) from 4 to have representatives in the committee: C(4,3) = 4 choices (one of them would be "A, B and C")

(2) The 3 couples chosen, choose the representative of each couple to be part of the committee: 2*2*2 = 8 choices (one of them would be "A1, B1 and C2")

From Multiplicative Principle we get 4*8 = 32.

Regards,
Fabio.

jimmy.student wrote:A committee of three people is to be chosen from four married couples. What is the number of differenct committees that can be chose if two people who are married to each other cannot both serve on the commitee?

16

24

26

30

32

OA E

this question was also discussed so many time before
there are 4 pairs of people => total =8 pepple.
the number of way to take out of 3 people in group of 8 is 8!/3!4!=56
there are 4 pairs, choose number of people in which there are a pair of husband and wife and 1 people from other pairs.
AA, BB, CC, DD
there are 3!*4 =24 combination ( AA can go with B1, or B2, C1, C2, D1,D2 = 6 combination * 4 pair=24)

What is the number of differenct committees that can be chose if two people who are married to each other cannot both serve on the commitee= total - number of poeple who are married to each other can be both on the committe.= 56-24=32