Combination problem

[BTGmoderatorRO]

A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?
A. 20
B. 22
C. 26
D. 30
E. 32
which other approach can be used to solve this question? why is C accurate than the rest? please solve
OA is e

[GMATGuruNY]

The posted problem is the same as the following:

The 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 serve on the committee?

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

Number of options for the 1st person = 8.
Number of options for the 2nd person = 6. (Of the 7 people left, we can't use the mate of the 1st person chosen, leaving 7-1= 6 choices.)
Number of options for the 3rd person = 4. (Of the 6 people left, we can't use the mates of the 2 people already chosen, leaving 6-2 = 4 choices.)

Since the ORDER of the selections doesn't matter -- ABC is the same COMMITTEE as BCA -- we divide by the number of ways to arrange the 3 people chosen:
(8*6*4)/3! = 32.

The correct answer is E.

One more approach:

Number of ways to choose 3 people from 8 options = 8C3 = (8*7*6)/(3*2*1) = 56.

Determine the probability that the selected committee does not include a married couple.
Let the 3 people selected be A, B and C.
A can be any of the 8 people.
The probability that B is not the spouse of A = 6/7. (Of the 7 remaining people, 6 are not married to A.)
The probability that C is not the spouse of A or B = 4/6. (Of the 6 remaining people, 4 are not married to A or B.)
Since we want both of these events to happen, we multiply the fractions:
6/7 * 4/6 = 4/7.

Thus:
Of the 56 possible committees, the number that do not include a married couple = (4/7) * 56 = 32.

The correct answer is E.

A third approach:

Select 3 couples:
From the 4 couples, the number of ways to choose 3 = 4C3 = (4*3*2)/(3*2*1) = 4.

From each of the 3 couples, select one member:
Number of options from the 1st couple = 2.
Number of options from the 2nd couple = 2.
Number of options from the 3rd couple = 2.
To combine these options, we multiply:
2*2*2 = 8.

To combine the options above, we multiply:
4*8 = 32.

The correct answer is E.

[Scott@TargetTestPrep]

A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?
A. 20
B. 22
C. 26
D. 30
E. 32
which other approach can be used to solve this question? why is C accurate than the rest? please solve
OA is e

Solution:

There are 8 options for the first member, 6 options for the second member (since the first member and his/her teammate cannot be selected) and 4 options for the third member.

If the order were important (for instance, if we were selecting a president, a vice president and a secretary); there would be 8 x 6 x 4 options. However, as the order is not important, there are (8 x 6 x 4)/3! = 8 x 4 = 32 possible selections.

Alternate Solution:

First, let’s select three teams out of four. Since there are four teams and we are selecting three, this can be done in 4 ways (just choose one team which is not to be selected).

Next, out of the three teams we pick, we have two options for each team (the first member or the second member). Thus, there are in total 4 x 2 x 2 x 2 = 32 possible selections.

Answer: E