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05 Aug 2025, 05:28
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scheol79
For Questions with small values like these and multiple conditions, I'd rather list down the cases than use permutations/combinations to avoid any mistakes.
This is how i'd exactly solve :
So here,
n=3
Total = 5
Jane(a), Joan(b), Paul(c), Stuart(d), and Jessica(e)
Paul(c) & Stuart(d) can't be together.
Jane(a) iff Paul(c)
Possibilities with a : {a,c,b} , {a,c,e} = 2
Possibilities with b except a : {b,c,e} , {b,d,e} = 2
Possibilities with c except a,b : -
rest all -
Thus ans is 4.
11 Apr 2026, 23:19
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Most effiicient way to solve is to write down all their names down and simply draw arrows with whom they can pair. Also make a distinction [according to ur convenience] of the pair that can't be formed.
Form teams without Jessica first and then with Jessica as she has added restrictions. This way one will not overlap cases is at all.
Without Jessica:
Paul and Stuart each getting into 1 team each with the other two Js.
2 teams.
With Jessica:
Has to go with Paul with only two options either Jane or Joan again as Paul can't go with Stuart.
2 teams.
Total = 4
Answer: Option B
Form teams without Jessica first and then with Jessica as she has added restrictions. This way one will not overlap cases is at all.
Without Jessica:
Paul and Stuart each getting into 1 team each with the other two Js.
2 teams.
With Jessica:
Has to go with Paul with only two options either Jane or Joan again as Paul can't go with Stuart.
2 teams.
Total = 4
Answer: Option B
Bunuel
12 Apr 2026, 02:31
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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?
Paul -------> x Stuart
Stuart -----> x Paul
x Paul -----> x Jane or Jane ------> Paul
Committee 1: Jane, Paul, Joan
Committee 2: Jane, Paul, Jessica
Committee 3: Paul, Joan, Jessica
Committee 4: Stuart, Joan, Jessica
IMO B
Paul -------> x Stuart
Stuart -----> x Paul
x Paul -----> x Jane or Jane ------> Paul
Committee 1: Jane, Paul, Joan
Committee 2: Jane, Paul, Jessica
Committee 3: Paul, Joan, Jessica
Committee 4: Stuart, Joan, Jessica
IMO B
12 Apr 2026, 10:11
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karishma B and @Bunnel
I did it like selecting 3 teams that can be formed out of 5 =>5C3 =10
Now considering the non possible options :
1. Paul Sturat Case: other1 can be chosedn in 3 ways hence =3 ways( P,S, Jane: P,S,Jone : P,S, Jessica )
2. Jane Paul Case: other one can be selected in 2 ways (considering overlap ) (Jane ,Paul, Jessica: Jane, Paul, Jone )
which gives 5 possible ways of not selecting
Ultimately giving us 5 ways of selection.
I would really request your advise on where i am missing out!!!
I did it like selecting 3 teams that can be formed out of 5 =>5C3 =10
Now considering the non possible options :
1. Paul Sturat Case: other1 can be chosedn in 3 ways hence =3 ways( P,S, Jane: P,S,Jone : P,S, Jessica )
2. Jane Paul Case: other one can be selected in 2 ways (considering overlap ) (Jane ,Paul, Jessica: Jane, Paul, Jone )
which gives 5 possible ways of not selecting
Ultimately giving us 5 ways of selection.
I would really request your advise on where i am missing out!!!
