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

OA later

for the sake of convenience lets name Jane=A, Joan=B, Paul= C, Stuart =D and Jessica= E

Combination possible are:

ABC, ACE, BDE, DEB. therefore 4 combinations. B

i hope i dint miss any.

Preet

singhpreet1 wrote:

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

OA later

for the sake of convenience lets name Jane=A, Joan=B, Paul= C, Stuart =D and Jessica= E

Combination possible are:

ABC, ACE, BDE, DEB. therefore 4 combinations. B

i hope i dint miss any.

Preet

You missed BCE. Answer should be 5

kvcpk wrote:

singhpreet1 wrote:

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

OA later

for the sake of convenience lets name Jane=A, Joan=B, Paul= C, Stuart =D and Jessica= E

Combination possible are:

ABC, ACE, BDE, DEB. therefore 4 combinations. B

i hope i dint miss any.

Preet

You missed BCE. Answer should be 5

Jane refuses to be in the committee without Paul

therefore BCE is not possible buddy.

Preet

selango 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?

jane=ja
joan=jo
paul=p
stuart=s
jessica=je
ja cant be with s
if p there then only ja will be there
now let select person on which there is no condition p,je will have to be 2 person,now e need to select third
p,je,ja
p,je,s
je,s,jo
p,jo,je

singhpreet1 wrote:

kvcpk wrote:

singhpreet1 wrote:

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

OA later

for the sake of convenience lets name Jane=A, Joan=B, Paul= C, Stuart =D and Jessica= E

Combination possible are:

ABC, ACE, BDE, DEB. therefore 4 combinations. B

i hope i dint miss any.

Preet

You missed BCE. Answer should be 5

Jane refuses to be in the committee without Paul

therefore BCE is not possible buddy.

Preet

Jane refuses to be in the committee without Paul
BUT
Paul doesnot refuse to be in the committee without Jane Buddy!!

Preet,

BDE and DEB are same option.

selango wrote:Preet,

BDE and DEB are same option.

yes! what is the OA?

P and S can never be together. If there is no P, then there is no Jane. Thus, if there is Jane, then there is P.

P__ __

Here, we can't have S (because we have P). We can have any two of the three "J"s though. So that's 3 combinations so far.



S__ __

Here, we can't have P (because we have S). Because we don't have P, we also don't have Jane. That leaves only the other two Js. So, this is 1 more combination.

Can we have a committee with neither of P and S (ie, all 3 Js)? No, we can't. The moment we don't have P, we can't have Jane, and that would only leave the 2 other Js, and so we can't make the three-person committee.

Choose B.

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

OA later

A very good question, which tests how imaginative we really are. Let me call the five candidates: Jane, Joan, Paul, Stuart, and Jessica as A, B, C, D, and E respectively, just for simplicity.

Now, a committee with C cannot have D to it, hence the remaining two could be selected from three candidates in 3C2 = 3 ways. Case won't be same if the committee is constructed with D, as it won't have C to it and so A would also walk out. Now, the remaining two could be selected from two candidates in 2C2 = 1 way. Total ways are [spoiler]4, therefore.

B[/spoiler]

sanju09,

wow...u use probability method to solve instead of brute force method :)I expected some probablity method.