blaster wrote:A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

with paul:
4
w/o paul:
4

8.

blaster wrote:A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

general possibilities =5C3=10
(I) W/O Stuart => Paul-Jane-Joan; Paul-Jane-Jessica; Paul-Joan-Jessica => 3 combination sets
(II) W/O Paul => Stuart-Joan-Jessica => 1 combination set
(III) W/O Paul and Stuart => Joan-Jessica-none => 0 combination set

Combining (I, II and III) => 1+3+0=4

blaster wrote:A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

what the answer?
night, 40 is impossible because there is only 5!/2!/3!=10 combinations to arrange 3 people in a group include P,jane and stuart

i consider paul, stuart, jane, joan, jess = 1,2,3,4,5
1 will not be with 2 and 3 so 1 just can be with 4 and 5
1 will not be with 2 so 2 will be with 3,4,5 .consider it 4 people in a group without 1 ( paul) the combination is 4!/3!=4 ( the combination can be 2,3,4 or 235 or 245 or 345)
my answer is 5
whats the OA?

diebeatsthegmat wrote:

blaster wrote:A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

what the answer?
night, 40 is impossible because there is only 5!/2!/3!=10 combinations to arrange 3 people in a group include P,jane and stuart

i consider paul, stuart, jane, joan, jess = 1,2,3,4,5
1 will not be with 2 and 3 so 1 just can be with 4 and 5
1 will not be with 2 so 2 will be with 3,4,5 .consider it 4 people in a group without 1 ( paul) the combination is 4!/3!=4 ( the combination can be 2,3,4 or 235 or 245 or 345)
my answer is 5
whats the OA?

Even did the same way and got the same answer as 5 , Just a small difference in my approach is was that i used A,B,C,D & E, for paul, stuart, jane, joan & jess respectively ..

Just wondering is there any other quick approach to solve this problem.

goyalsau wrote:

diebeatsthegmat wrote:

blaster wrote:A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

what the answer?
night, 40 is impossible because there is only 5!/2!/3!=10 combinations to arrange 3 people in a group include P,jane and stuart

i consider paul, stuart, jane, joan, jess = 1,2,3,4,5
1 will not be with 2 and 3 so 1 just can be with 4 and 5
1 will not be with 2 so 2 will be with 3,4,5 .consider it 4 people in a group without 1 ( paul) the combination is 4!/3!=4 ( the combination can be 2,3,4 or 235 or 245 or 345)
my answer is 5
whats the OA?

Even did the same way and got the same answer as 5 , Just a small difference in my approach is was that i used A,B,C,D & E, for paul, stuart, jane, joan & jess respectively ..

Just wondering is there any other quick approach to solve this problem.

I corrected the solution, as the approach was mechanical; this is simple combination with elimination

I end with 4 sets though

blaster wrote:A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

According to me the answer is 4 teams
Plz advise whether it is correct.

The approach is as below.
Lets start with Jane

1) If Jane is there in the committee, even Paul has to be there.
Jane+Paul+third member
Third member cannot be Stuart.So the third member can be either Joan or Jessica
So the two two teams can be as below
Jane+Paul+Jessica
Jane+Paul+Joan

2) If we start forming a community with Joan, then
a) if we add Jane we will have to add hence same as 1)
b) if we add Paul then we have then we have two options for the third member--Jane/Jessica.we can't take stuart
We have already considered Jane in 1) Hence the new team can be
Joan+Paul+Jessica
c) if we add stuart then we have then we have only one option for the third member i.e Jessica.we can't take stuart. He the new team can be
Joan+Staurt+Jessica

Repeating the same procedure for remaining three members we come to the conclusion that no more new teams apart from the above 4 can be formed.
Hence the answer is 4.