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# M10-25

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Math Expert
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M10-25  [#permalink]

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15 Sep 2014, 23:42
00:00

Difficulty:

95% (hard)

Question Stats:

40% (01:48) correct 60% (01:46) wrong based on 119 sessions

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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

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Posts: 51100
Re M10-25  [#permalink]

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15 Sep 2014, 23:42
Official Solution:

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

# of committees with Paul is $$C^2_3=3$$, since we choose remaining 2 members out of 3 people: Jane, Joan and Jessica (so everyone but Stuart);

# of committees with Stuart is $$C^2_2=1$$, since we choose remaining 2 members out of 2 people: Joan and Jessica (so everyone but Paul and Jane);

Total: $$3+1=4$$. Notice that no committees are possible if neither Paul nor Stuart are there, since no Paul means no Jane too, so only 2 members are left (Joan and Jessica) which cannot form 3-member committee.

Answer: B
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Re: M10-25  [#permalink]

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31 Mar 2016, 09:50
Correct me if I'm mistaken;
in the solution it should be 1 committee with Paul and 3 with Stuart.
thanks
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Re: M10-25  [#permalink]

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31 Mar 2016, 10:34
Avigano wrote:
Correct me if I'm mistaken;
in the solution it should be 1 committee with Paul and 3 with Stuart.
thanks

Can you please quote which part of the solution do you refer to? Thanks.
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Re: M10-25  [#permalink]

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31 Mar 2016, 11:17
Quote:
# of committees with Paul is C 2 3 =3
C32=3
, since we choose remaining 2 members out of 3 people: Jane, Joan and Jessica (so everyone but Stuart);

# of committees with Stuart is C 2 2 =1
C22=1
, since we choose remaining 2 members out of 2 people: Joan and Jessica (so everyone but Paul and Jane);

Isn't it reversed? Paul 1 and Stuart 3?

Thank you
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Joined: 02 Sep 2009
Posts: 51100
Re: M10-25  [#permalink]

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31 Mar 2016, 11:27
2
Avigano wrote:
Quote:
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?

# of committees with Paul is C 2 3 =3
C32=3
, since we choose remaining 2 members out of 3 people: Jane, Joan and Jessica (so everyone but Stuart);

# of committees with Stuart is C 2 2 =1
C22=1
, since we choose remaining 2 members out of 2 people: Joan and Jessica (so everyone but Paul and Jane);

Isn't it reversed? Paul 1 and Stuart 3?

Thank you

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

{Stuart, Joan and Jessica}
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Re: M10-25  [#permalink]

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31 Mar 2016, 11:36
I just realized that I read "Jane refuses to be in the committee without Paul".. as "Jane refuses to be in the committee WITH Paul"
So yeah, well Sh"#\$ happens! My ugly mistake, pardon me.
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Re: M10-25  [#permalink]

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20 May 2016, 05:12
Bunuel wrote:
Official Solution:

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

# of committees with Paul is $$C^2_3=3$$, since we choose remaining 2 members out of 3 people: Jane, Joan and Jessica (so everyone but Stuart);

# of committees with Stuart is $$C^2_2=1$$, since we choose remaining 2 members out of 2 people: Joan and Jessica (so everyone but Paul and Jane);

Total: $$3+1=4$$. Notice that no committees are possible if neither Paul nor Stuart are there, since no Paul means no Jane too, so only 2 members are left (Joan and Jessica) which cannot form 3-member committee.

Answer: B

Hi Bunnel,

I have worked as below :

With paul we have total 4 members to form 3 students as committe i.e 4C3 = 4C1 = 4
with stuart we have only combination , 3C3 =1

total = 4+1 = 5

could you please help me where am i going wrong.

Thanks.
Math Expert
Joined: 02 Aug 2009
Posts: 7099
Re: M10-25  [#permalink]

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20 May 2016, 05:43
babuvgmat wrote:
Bunuel wrote:
Official Solution:

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

# of committees with Paul is $$C^2_3=3$$, since we choose remaining 2 members out of 3 people: Jane, Joan and Jessica (so everyone but Stuart);

# of committees with Stuart is $$C^2_2=1$$, since we choose remaining 2 members out of 2 people: Joan and Jessica (so everyone but Paul and Jane);

Total: $$3+1=4$$. Notice that no committees are possible if neither Paul nor Stuart are there, since no Paul means no Jane too, so only 2 members are left (Joan and Jessica) which cannot form 3-member committee.

Answer: B

Hi Bunnel,

I have worked as below :

With paul we have total 4 members to form 3 students as committe i.e 4C3 = 4C1 = 4
with stuart we have only combination , 3C3 =1

total = 4+1 = 5

could you please help me where am i going wrong.

Thanks.

You are going wrong in the highlighted portion...
If paul is there in the team, stuart cannot be there and we have to choose 2 out of remaining three = 3C2 = 3..
If paul is not there, jane will also be not there, there is ONLY one team possible 1C1=1..

total 3+1=4..

whereas in the highlighted portion if you take 4C3, you are adding one scenario where paul is not there and Jane is there
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: M10-25  [#permalink]

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20 May 2016, 06:33
whereas in the highlighted portion if you take 4C3, you are adding one scenario where paul is not there and Jane is there[/quote]

Ohh yeah .. this point i was missing .

Thank you Bunnel
Math Expert
Joined: 02 Aug 2009
Posts: 7099
Re: M10-25  [#permalink]

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20 May 2016, 06:46
1
babuvgmat wrote:
whereas in the highlighted portion if you take 4C3, you are adding one scenario where paul is not there and Jane is there

Ohh yeah .. this point i was missing .

Thank you Bunnel [/quote]

hi,
Bunuel, you have so many great posts floating around that any post clearing any query is believed to be yours .....
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Joined: 16 Mar 2016
Posts: 129
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M10-25  [#permalink]

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22 May 2016, 07:11
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?

I have worked in a different way :

1/ The total number of committees : 3 from 5 = $$\frac{(5*4*3)}{(3*2*1)}$$= 10

2/ The number of committees which break the rules :
i.e. : the number of committees with Paul and Stuart, and the number of committees with Jane and not Paul

a) with Paul and Stuart : 1*1*3 = 3 committees

b) with Jane and not Paul : $$\frac{(1*3*2)}{(2*1)}$$ = 3 committees as well

Finally, 10 - 3 -3 = 4
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Re: M10-25  [#permalink]

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30 May 2016, 18:18
Alex75PAris 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?

I have worked in a different way :

1/ The total number of committees : 3 from 5 = $$\frac{(5*4*3)}{(3*2*1)}$$= 10

2/ The number of committees which break the rules :
i.e. : the number of committees with Paul and Stuart, and the number of committees with Jane and not Paul

a) with Paul and Stuart : 1*1*3 = 3 committees

b) with Jane and not Paul : $$\frac{(1*3*2)}{(2*1)}$$ = 3 committees as well

Finally, 10 - 3 -3 = 4

Hi,

Why did you divided 2*1 into 1*3*2 when you were computing numbers around Jane without Paul?
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Re: M10-25  [#permalink]

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06 Jul 2016, 04:43
Another approach:

Committees with Jane >> Paul is member, Stuart excluded >> Committee Possibilities = Possibilities for 3rd member: 2 (either Joan or Jessica)

Committees without Jane >> Paul cannot be with Stuart (enemies restriction) >> treat Paul + Stuart as 1 unit (ways to pick within unit = 2), need to pick 3 from 3 choices
>> Committee Possibilities = (3!/3!) * 2 = 2

Total committee possibilities = 2 + 2 = 4
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Re: M10-25  [#permalink]

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02 Sep 2016, 09:54
Jane, Joan, Paul, Stuart, and Jessica
jane and paul together ( if paul is there stuart is not there) so out of remaining three we can choose only 02- (jane, paul, jessica) or (jane,paul, joan)

if paul is not there jane is not there so 01 combination - (joan, stuart, jessica)

if paul is there stuart can not be there, also jane dosnt want to be without paul, she can be outside of the group too ( paul, jesscia, joan

total =4
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Re M10-25  [#permalink]

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05 Apr 2018, 09:21
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.
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Re: M10-25  [#permalink]

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05 Apr 2018, 12:43
nilotpal1990 wrote:
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.

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

{Stuart, Joan and Jessica}
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Re: M10-25  [#permalink]

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04 Aug 2018, 01:10
A really good question! Took awhile to solve but nevertheless got it finally.

Total ways = $$5C3$$ = 10
P(Paul + Stuart) = $$3C1 * 2C2 = 3$$
P(Jane without Paul) = $$3C2 * \frac{2C1}{2} = 3$$

Answer is 10-3-3 = 6
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Re: M10-25  [#permalink]

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26 Aug 2018, 04:30
hi bunuel,
I solved it this way:
if jane is on the committee then paul will be on committee so the last place can be filled in 2 ways.
if stuart is on committee then we have to fill the remaining two places with joan and jessica in 2*1 ways.So total ways are 2+2=4 ways.
Is this method correct?
Re: M10-25 &nbs [#permalink] 26 Aug 2018, 04:30
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# M10-25

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