Married couple - Combo

[susan.meng]

A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

a. 16
b. 24
c. 26
d. 30
e. 32

[Ben Ku]

Hi Susan,

Please explain the specific question you have about the problem. What have you tried? Where are you getting stuck?

As a hint, we need to first select three couples. Then we need all combinations of husbands and wives from these three couples.

[tigerwoods]

the answer should be 24

a1a2
b1b2
c1c2
d1d2

these are four couples.....

we take 1*3*2= 6 possible for one person from one couple for example

1st 2nd 3rd
a1 __ either(b or c or d) _ either (of the two not selected in the second spot)

hence ( 1*3*2 ) *4 = 24........

please suggest.....

[cyber_office]

I think the answer should be E. First solve the problem as though there is no restriction:

Choose 3 people from 8 people (where order does not matter)

8C3 = (8 * 7 * 6 * 5!)/(3!)*(5!) = 56

56 includes all combinations including those in which husband and wife are on the committee.

Since we have 4 married couples we can have each of them serve on a committee and third place could be filled by one of the other 6 remaining people. For example, if A1 and A2 are husband and wife, the arrangement would look like:

A1 A2 (3rd slot - any of the other six)

Similarly,

B1 B2 (3rd slot - any of the other six, which in this case includes A1 and A2)

Therefore, we have 4 possibilites (for husband and wife combo) * 6 of the remaining 8

=> 4*6 = 24 combinations have husband-wife on committee.

Therefore, 56 - 24 = 32 will satisfy the requirement that no husband-wife combo is together on a committee.

[RonPurewal]

A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

a. 16
b. 24
c. 26
d. 30
e. 32

the fastest way here is to use the SLOT METHOD. (combinatorics formulas are walking death on this problem; they pretty much just don't work, because of the restrictions imposed in the problem.)

as usual, the slot method has three steps.

1 * CREATE THE SLOTS
you have three decisions to make (3 members on the committee), so there are three slots.

2 * FILL IN THE SLOTS WITH THE #'S OF OPTIONS
for the first slot, there are 8 options (you can pick anyone).
for the second slot, there are 6 options (you can't pick the first person, and, per the restriction, you can't pick that person's spouse).
for the third slot, there are 4 options (you can't pick either of the first two people, OR either of their spouses).
-- so, the slots are filled in with 8, 6, 4.

3 * DIVIDE BY THE FACTORIAL IF ORDER DOESN'T MATTER
here, order doesn't matter, since you're just picking "a committee" (with no distinctions between the three positions on the committee). so divide by 3!.

so the answer is
(8 x 6 x 4) / 3!
= 8 x 4 (since 3! = 6, and so those two terms cancel)
= 32.

ans (e)