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A committee of 3 students has to be formed. The are five [#permalink]
28 Jul 2007, 11:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
A committee of 3 students has to be formed. The 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 combinations are possible ?
A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3 (ii) 4 (iii) 5 (iv) 6 (v) 8
Should be (ii).
If Paul is in the committee there can be 3C2 = 3 committees. (Stuart won't be there.)
I Stuart is in the committee there can only be 1 committee. (Paul and Jane won't be there.)
There can be no committee without either of them. (Stuart, Paul and Jane won't be there, so there are only 2 left.)
A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3 (ii) 4 (iii) 5 (iv) 6 (v) 8
there are 5C3 ways to make a comittee of 3 students out of 5 w/o restriction. = 10
there are 3 ways to have a team with Paul and Stuart
there are 3 ways to have a team with Jane and Paul
There is 1 way to have a team with Jane Paul and Stewart, but it is counted twice above.
A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3 (ii) 4 (iii) 5 (iv) 6 (v) 8
A. i would say 3.
(jean + Paul) + (either joan or jessica) = 2
(paul) + (joan and jessica) (i.e. not stuart and jean) = 1
A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3 (ii) 4 (iii) 5 (iv) 6 (v) 8
A. i would say 3.
(jean + Paul) + (either joan or jessica) = 2 (paul) + (joan and jessica) (i.e. not stuart and jean) = 1
so total = 3
i didn't do this the formal way, but there are 4 groups
S A B
P J A
P J B
P A B
it states jean needs to be with paul, but not the other way around
A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3 (ii) 4 (iii) 5 (iv) 6 (v) 8
A. i would say 3.
(jean + Paul) + (either joan or jessica) = 2 (paul) + (joan and jessica) (i.e. not stuart and jean) = 1
so total = 3
i didn't do this the formal way, but there are 4 groups
S A B P J A P J B P A B
it states jean needs to be with paul, but not the other way around
you are right.
i also got 4 when i used formuala but thought should use this approach to minimize the risk. hmmmmmmm fell on the trap. toooo sad!!!!!!
A committee of 3 students has to be formed. The 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 combinations are possible ?
(i) 3 (ii) 4 (iii) 5 (iv) 6 (v) 8
A. i would say 3.
(jean + Paul) + (either joan or jessica) = 2 (paul) + (joan and jessica) (i.e. not stuart and jean) = 1