M10-25

Question

A committee of three students has to be formed from five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be on the committee together, and Jane refuses to be on the committee without Paul, how many committees can be formed?

A. 3
B. 4
C. 5
D. 6
E. 8

A. 3
B. 4
C. 5
D. 6
E. 8

Bunuel · Accepted Answer

Official Solution:

A committee of three students has to be formed from five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be on the committee together, and Jane refuses to be on the committee without Paul, how many committees can be formed?

A. 3
B. 4
C. 5
D. 6
E. 8

A. 3
B. 4
C. 5
D. 6
E. 8

First, we note that a committee cannot be formed if neither Paul nor Stuart is on the committee. This is because, without Paul, Jane also refuses to be on the committee, leaving only two other candidates (Joan and Jessica), which is not enough to form a three-member committee.
 
Now, let's consider the committees that include Paul. Since Stuart refuses to be on the committee with Paul, we must choose two other members from the remaining candidates: Jane, Joan, and Jessica. This gives us  C^2_3=3  possible committees.
 
Similarly, we can count the committees that include Stuart but not Paul. Since Jane refuses to be on the committee without Paul, we must choose two other members from the remaining candidates: Joan and Jessica. This gives us  C^2_2=1  possible committee.
 
Thus, the total number of possible committees is  3+1=4 .

Answer: B

nilotpal1990 · Answer

I think this is a  poor-quality  question and I don't agree with the explanation. when paul is a part of a committee then jane has to be part of that committee.so no of possible cases should be : 
paul,jane,jessica/joan.
again when stuart is a part of that committee jane & paul wont be a part of that.so no of possible case is 1.
so the total no of possible cases are :3.

Bunuel · Answer

It's not that hard to list the committees:
{Paul, Jane, Joan}
{Paul, Jane, Jessica}
{Paul, Joan, Jessica}

{Stuart, Joan and Jessica}

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Zaf1997 · Answer

Hi there,

Do I understand correctly that in the explanation Jane and Paul are in the same committee? If this is correct, would it not contradict the condition given by the question, that Jane refuses to be in the team with Paul?

Now, let's consider the committees that include Paul. Since Stuart refuses to be on the committee with Paul, we must choose two other members from the remaining candidates: Jane, Joan, and Jessica. This gives us 3C2=3

Bunuel · Answer

The question says:  ...Jane refuses to be on the committee  without  Paul... , which means that if Jane is on the committee, then Paul must also be on it. Here are possible committees:

{ Paul, Jane,  Joan}
{ Paul, Jane,  Jessica}
{Paul, Joan, Jessica}

{Stuart, Joan and Jessica}

Anugmat740 · Answer

Total Case  5c3 =10

Cases including Paul and Stuart= 3
{Paul, Stuart, Jane} {Paul, Stuart,Joan}, {Paul, Stuart, Jessica}

Cases including Jane without paul = 3
{Jane, Joan, Stuart}, {Jane, Joan, Jessica}, {Jane, Stuart, Jessica}

Required Cases =10 -6 =4

A_Nishith · Answer

Total Case 5c3 =10

Cases including Paul and Stuart= 3 (i.e. 3C1 -- as 3 member to fill 1 position)
{Paul, Stuart, Jane} {Paul, Stuart,Joan}, {Paul, Stuart, Jessica}

Cases including Jane without paul = 3 (i.e 3C2 --- as 3 person left to fill 3 position & order does not matter in Team Making)
{Jane, Joan, Stuart}, {Jane, Joan, Jessica}, {Jane, Stuart, Jessica}

Required Cases =10 -6 =4

rak08 · Answer

This is a "Poor Quality" Ques.

We can have

Jane , Paul , Jessica
Stuart, Jane,Joan
Stuart, Jane, Jessica
Stuart, Paul, Joan
Joan, Paul,Jessica
Stuart, Jessica, Joan

6 cases mentioned above

Whats wrong??  Bunuel

Bunuel · Answer

The question is fine! You should read the question, solution, and discussion more carefully.

Jane can't be on the committee without Paul 
 Paul and Stuart can't be together

Among your 6 cases, 3 violate these rules, so only 3 are valid. One more valid case is missing from your list, making the correct total 4.

The question says:  ...Jane refuses to be on the committee  without  Paul... , which means that if Jane is on the committee, then Paul must also be on it. Here are possible committees:

{ Paul, Jane,  Joan}
{ Paul, Jane,  Jessica}
{Paul, Joan, Jessica}

{Stuart, Joan and Jessica}

Paattaa · Answer

I like the solution - it’s helpful.

maztho · Answer

I like the solution - it’s helpful. total = 5C3 = 10

Subtract Paul and Stuart together:
P S _ (3 choices) = 3

Subtract Jane and 'Paul
J _ _ (3C2 = 3) 3C2 not 4C2 because Paul is off

10 -3 - 3 = 4

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