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A woman has 11 close friends. Find the number of ways she
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12 Jun 2005, 08:42
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A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner, (1) but where 2 of the friends are married and will not attend separately; (2) but where 2 of the friends are not on speaking terms and will not attend together. == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



Senior Manager
Joined: 30 May 2005
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Case 1
Total # of cases = N(couple invited) + N(Couple not invited)
=C(9,3)*C(2,2) + C(9,5)
=84+126
=210
Case 2
=All possible combos  Combos with both fighting friends
=C(11,5)  C(9,3)*C(2,2)
=46284
=378



Manager
Joined: 07 Mar 2005
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way above my head. Is there a good website for probability help?



Director
Joined: 18 Apr 2005
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Re: A woman! PS
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12 Jun 2005, 13:17
Dan wrote: A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner, (1) but where 2 of the friends are married and will not attend separately; (2) but where 2 of the friends are not on speaking terms and will not attend together.
1) 7C5 + 7 + 2*7C2 =7(3+1 +6) = 70
2) not very specifc. are the freinds who don't talk married or not?



Director
Joined: 18 Apr 2005
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Location: Canuckland

tkirk32 wrote: way above my head. Is there a good website for probability help?
this problem is about counting.
here is what you need to remember
# of ways of pairing n ppl with m ppl = n*m
# of ways to arrange n different objects = n!, if it's around the table it's (n1)!
if among n objects there are, for example, 4 groups of identical objects with # of object being x1,x2,x3, and x4 then it is n!/x1!x2!x3!x4!
# of ways to group n different objects = 2^n, 2 is 2 possibilities  included and not included
# of ways to pick n objects from m different objects (permutations) when order of n matters
is mPn = m!/(mn)!
# of ways to pick n objects from m different when order of picked objects doesn't matter (combinations) mCn = m!/(mn)!n!
I was a counting newbie too before I joined the forum, then I tried to solve few problems on the forum and understood how counting worked.
Just make sure you understand the logic behind every formula, if you do you are set. and the logic is quite simple.



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Re: A woman! PS
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12 Jun 2005, 13:51
sparky wrote: Dan wrote: A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner, (1) but where 2 of the friends are married and will not attend separately; (2) but where 2 of the friends are not on speaking terms and will not attend together. 1) 7C5 + 7 + 2*7C2 =7(3+1 +6) = 70 2) not very specifc. are the freinds who don't talk married or not?
Can you pls explain? How did you get the 7  She has 11 friends and in either case, only 2 of them have peculiarities.
Thanks



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Re: A woman! PS
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12 Jun 2005, 13:59
AJB77 wrote: sparky wrote: Dan wrote: A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner, (1) but where 2 of the friends are married and will not attend separately; (2) but where 2 of the friends are not on speaking terms and will not attend together. 1) 7C5 + 7 + 2*7C2 =7(3+1 +6) = 70 2) not very specifc. are the freinds who don't talk married or not? Can you pls explain? How did you get the 7  She has 11 friends and in either case, only 2 of them have peculiarities. Thanks
I did 1) assuming that 2 friends are married (the problem should have read 'marrierd to each other' to be precise not just married), meaning there are 2 couples
7C5  # of ways to invite only single firends
7 is # of ways to to invite 2 couples + 1 extra single friend
2*7C2  # of ways to invite 1 couple and 3 single friends
if the problem in 1) means only 1 couple
then it is 9C5 + 9C3 = 7*30 = 210 (same thing you got )



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sparky,
Thanks, can you help jog my memory? How did you reduce C(9,5) + C(9,3) to 30*7 ?
Thanks



Director
Joined: 18 Apr 2005
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AJB77 wrote: sparky,
Thanks, can you help jog my memory? How did you reduce C(9,5) + C(9,3) to 30*7 ?
Thanks
pull out 7 since it is common and cannot be divided by anything, and do the rest of the calculations



Senior Manager
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I am assuming here that the two cases are independent...though i could be wrong.
Case 1:
If neither of the married friends is invited, the number of ways to choose 5 people out of the remaining 9 is 9C5
If one of the married friends is invited, the num of ways to choose 3 more people to call is 9C3 (two possible combinatios as either married friend could be invited...thus 9C3*2)
If both married friends are invited, the num of ways to choose the last person to call is 9C1
Thus answer = 9C5 + 9C3*2 + 9C1
Case 2:
If neither of the people who dont get along are invited, the number of ways to choose 5 people is 9C5
If one of them is invited, the way to choose remaining 4 is 9C4 (two ways to do this, as either of the two could be chosen...thus 9C4*2)
Thus, answer = 9C5 + 9C4*2
If both cases are meant to be combined and it is assumed that the people who dont get along are not the married ones:
If none of the exceptions chosen: 7C5
If one married couple chosen and 3 normal: 7C3 (*2)
If both married and 1 normal: 7C1
If one married couple, 1 angry person and 2 normal: 7C2 (*2*2)...as there are two possibilities of choosing a married couple and 2 of choosing an angry person.
If both married people and 1 angry: 2 (2 ways to choose the angry person)
If 1 angry person and 4 normal people chosen: 7C4*2 (as either angry person cuold be chosen)
Thus ans: 7C5+7C3*2+7C1+7C2*4+2+7C4*2
no idea if any of that made sense
and sorry about the strange naming ('angry' 'normal')



Manager
Joined: 28 Aug 2004
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1) married to each other..so it is one couple.
2) is indep of 1), so nothing about being married.
OA is 210 and 252 resp.



Senior Manager
Joined: 30 May 2005
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Dan,
How did you get your second answer? Seems to me that you are only counting the cases when EITHER one of the fighting friends is selected but you did not count the cases where NEITHER is selected.
cloudz9 and I got the same answer 378



Manager
Joined: 28 Aug 2004
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AJB77 wrote: Dan,
How did you get your second answer? Seems to me that you are only counting the cases when EITHER one of the fighting friends is selected but you did not count the cases where NEITHER is selected.
cloudz9 and I got the same answer 378
am getting 11C5  9C3 = 378, but the OA (official answer) is 252.



Senior Manager
Joined: 30 May 2005
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Dan,
Where did you find this question? can you please post it verbatim when you get a minute?
Thanks



Manager
Joined: 05 May 2005
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Location: Kyiv, Ukraine

could it be 10C5?
10!/5!5!=10*9*8*7*6/5*4*3*2=252
either one of the 2 people who are fighting is not counted, so we can pick from 10 for 5 places



Manager
Joined: 28 Aug 2004
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july05 wrote: could it be 10C5?
10!/5!5!=10*9*8*7*6/5*4*3*2=252
either one of the 2 people who are fighting is not counted, so we can pick from 10 for 5 places
interesting! but how is this different from 11C5  9C3?
I think you're assuming that 1 specific person is not invited  a restriction which is not stated/implied in the Q stem.



Director
Joined: 13 Nov 2003
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to AJB77 , think that for case 2 you have assumed that none of the fighting people is invited and thus you have 9C5 more cases. IMO case 2 has always one of the fighting friends invited which yields 2x9C4 or 252



Senior Manager
Joined: 30 May 2005
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BG wrote: to AJB77 , think that for case 2 you have assumed that none of the fighting people is invited and thus you have 9C5 more cases. IMO case 2 has always one of the fighting friends invited which yields 2x9C4 or 252 Okay. but the question just says that the 2 fighting friends should not attend together. It does not say that one of them must always attend. I'd have thought that ETS would not leave room for interpretation in Math. It would be interesting to know if both 252 and 378 were choices.



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Re: A woman! PS
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15 Jun 2005, 21:28
Dan wrote: A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner, (1) but where 2 of the friends are married and will not attend separately; (2) but where 2 of the friends are not on speaking terms and will not attend together.
(1) Choose the couple C(9,3)
Not choose the couple C(9,5)
C(9,3)+C(9,5)=9*8*7/6+9*8*7*6/24=9*8*7*(1/6+1/4)=210
(2) Choose one friend C(9,4)
Choose another C(9,4)
Choose neither C(9,5)
2C(10,4)+C(9,5)=2*9*8*7*6/24+9*8*7*6/24=9*8*7*6*(1/12+1/24)=9*7*6=378
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Intern
Joined: 15 Jun 2005
Posts: 10

A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner, (1) but where 2 of the friends are married and will not attend separately; (2) but where 2 of the friends are not on speaking terms and will not attend together.
1) Couple are invited + couple are not invited
9C3+ 9C5
84 + 126
210
2) If two friends don't attend together we have only ten friends left.
10c5
252







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