A committee of three people is to be chosen from four teams

Question

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

Bunuel · Accepted Answer

If you want to solve with slot method the simpler solution would be: 8*6*4/3!=32, 8 ways to choose for the first slot, 6 ways to choose for the second slot, 4 ways to choose for the third slot and dividing by 3! to get rid of duplication.

Another way to solve this problem would be: \(C^3_4*2^3=32\), where \(C^3_4\) is # of ways to choose which 3 team members out of 4 will be represented in the committee and multiplying this by 2*2*2 since we can choose any of 2 members from each chosen team.

Answer: E.

Check some VERY similar question to practice:
https://gmatclub.com/forum/if-a-committ ... 98533.html
https://gmatclub.com/forum/ps-combinations-94068.html
https://gmatclub.com/forum/ps-combinations-101784.html
https://gmatclub.com/forum/committee-of-88772.html
https://gmatclub.com/forum/if-4-people- ... 99055.html
https://gmatclub.com/forum/if-there-are ... 99992.html
https://gmatclub.com/forum/a-committee- ... 94068.html

Hope it helps.

Arthur1000 · Answer

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

My Method

First we find out the amount of combinations of 3 person teams from a pool of 8. C(3/8) = 8!/5!*3! = 56 ways

Next I find the number of ways we CAN make a 3 person team using two from the same group, so

We take 2 people from a 2 person group C(2/2) and mutiply that by taking any 1 person from the remaining 6 C(1/6)

C(2/2)*C(1/6) = 6

This is the number of combinations by taking both parties from pair A. As we have a four pairs we must multiply this by 4 (4x6 = 24)

So there are 24 ways in which two people from the same pair can work together.

Finally we subtract this from the total number of combinations to find the number of groups where pairs do NOT work together....

56 - 24 = 32!

Archit143 · Answer

Hi Brunel

I did not inderstand the concept behind writing 4c3 it means selecting 3 person out of 4 but it is 4 team so it should be 8c3

Am i Correct

Regards]

Archit

Bunuel · Answer

4C3 is selecting 3  teams  out of four. Then, since  each  team can give either of its two members for the committee, we should multiply 4C3 by 2*2*2.

Hope it's clear.

vsprakash2003 · Answer

Bunuel - What do you mean by selecting "3 teams" out of 4? The question states that 3 people form a committee (or team) and need to be selected from 4 teams of 2 people each. Archit is right. Shouldn't it be 8C3?

However the slotting method you mention is correct. 8*6*4/3!

However the slotting method you mention is correct. 8*6*4/3!

Bunuel · Answer

There are 4 teams. Now, if we select 3 teams from those 4 and each will send one member then thee committee will have 3 members and no 2 members from the same team.

Hope it's clear.

P.S. Please follow the links in my post above for similar questions to practice.

Archit143 · Answer

i think it should be only 4c3 * 2c1
3 teams out of four and 1 person out of 2
am i correct or missing sthn

Bunuel · Answer

EACH  team out of 3 chosen can send  ANY  of its 2 members, so its 2*2*2*4C3.

bapun11 · Answer

in the slot method why we divide by 3! to get rid of duplication? I am not sure about this concept. it would be great if any one explain

ENGRTOMBA2018 · Answer

It is an arrangement question similar to finding arrangements of AABBB = 5!/ (2!*3!) where 2! and 3! are done to remove the duplication of A and B as all As and all Bs are the same. Had all As or all Bs be different (A1 A2) or (B1 B2 B3) , then the answer would have been = 5!

Hope this helps.

Sheersh · Answer

Here is how I solved this -

A/A1, B/B1, C/C1, D/D1

Total ways of choosing without any ristriction: 8C3 = 8!/3!5! = 56

Now total number of cases that are not required -

1. One couple is picked - 2C2
2. Remaining one member is picked from remaining 6 people - 6C1
3. Since there are 4 couple so this can be multiplied by 4.

total: 4*2C2*6C1 = 24

Hence, total number desired : 56-24 = 32

13S12 · Answer

How I Solved it:

Teams with no couple chosen = Total no. of possible teams - Teams with one couple chosen
 = 8C3 - 4C1x6C1
 = 56 - 24 = 32

where:
4C1 --> From 4 couples choose any one
6C1 --> From the 6 remaining candidates, choose any one

AnisMURR · Answer

Hi guys,

I have prepared a video with detailed explanation of this question.

Click here to see the explanation.

ScottTargetTestPrep · Answer

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

swatib28 · Answer

I have solved this question in this way, Lets say we have A, B, C and D teams with 2 members each. Our goal is to select 3 out of 4 which can be done in 4 ways ( ABC,BCD,CDA,DAB) . Now,lets take one way ABC, each team A or B or C have 2 members each

(1+1)--> a team (1+1)--> B team (1+1)--> C team

Again , all teams has to send so will multiple.

2 * 2 *2 --> 8 * 4(ways)--> 32 Answer.

(1+1)--> a team (1+1)--> B team (1+1)--> C team

Again , all teams has to send so will multiple.

2 * 2 *2 --> 8 * 4(ways)--> 32 Answer.

AbhishekAnil · Answer

It will be 4C3 *2 *2 *2 
4C3 - represents the selection of 3 teams, 
2*2 *2 represnt the selection of either one of the 2 team members from each of the 3 selected teams

A   chit143

seyma · Answer

Hi Bunuel, I understood your method the best, but I am still not getting the 3! part. Could you please explain it?

Bunuel · Answer

8 * 6 * 4 counts the same three people in different orders. But in a committee the order does not matter, so each group of three people is counted 3! = 6 times. Therefore we divide by 3! to remove the duplicates.

P.S. Please follow the links in my post above for similar questions to practice.

A committee of three people is to be chosen from four teams

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A committee of three people is to be chosen from four teams

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