Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A woman has 11 close friends. Find the number of ways she [#permalink]
12 Jun 2005, 07:42

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.

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?

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 (n-1)!
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!/(m-n)!
# of ways to pick n objects from m different when order of picked objects doesn't matter (combinations) mCn = m!/(m-n)!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.

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.

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 )

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')

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.

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.

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

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.

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 _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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