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:

In a room filled with 7 people, 4 people have exactly 1 [#permalink]

Show Tags

27 Sep 2006, 09:31

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

In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

If you draw graph of 7 items ( form connections between them), you will find that there is essentially only one way to allocate (connections) such that some have 2 and some have one. In other words, there only one number of people that can have one friend.

Hope this helps.

I remember this being on the MGMAT tests. I wanted to point out that if you're studying prob./comb/perm, know the formulae, but don't depend on them. The GMAT is very likely to throw you a question that cannot be solved completely by the use of formulae.

Ok for this question to make sense we have to consider the permutations not combinations, I mean if A is friend with B then that does not mean B is friend with A (and vice versa) thus for people A, B, C, D, E, F

if you put them on 7 vertices of a polygon with 7 sides and AB is not same as BA, then there are 42 total pairs

a has 1 connection (b)
b has 2 connections (a,c)
c has 2 connections (b,d)
d has 1 connection (c)
e has 1 connection (f)
f has 2 connections (e,g)
g has 1 connection (f)

combinations including a: cdefg = 5
b: defg=4
c: efg = 4 (a considered earlier)
d: efg = 3 (a, b considered earlier)
e: g = 1

Ok for this question to make sense we have to consider the permutations not combinations, I mean if A is friend with B then that does not mean B is friend with A (and vice versa) thus for people A, B, C, D, E, F

if you put them on 7 vertices of a polygon with 7 sides and AB is not same as BA, then there are 42 total pairs

Of these 10 result in friends (3 X2 + 4 X 1)

Thus remaining = 32

probability = 32/32 = 16/21

Even though I also got 16/21, my approach is different

I believe the question implies that if A is a friend of B then B is a friend of A, thus these two have exactly one friend. With this approach we can split the 7 people into two groups - ABCD and EFG, where A and B, C and D, E and F and G are friends.

Now, there are 21 (7C2) outcomes as order doesn't matter. Now we want to count favorable outcomes:

For A - 5 (AC, AD, AE, AF, AG) - A can't be chosen with B
For B - 5 (BC, BD, BE, BF, BG) - B can't be chosen with A
For C - 3 (CE, CF, CG) - C has been chosen with A and B and can't be chosen with D
For D - 3 (DE, DF, DG) - D has been chosen with A and B and can't be chosen with C

E, F and G can't be chosen with each other and have already been chosen with ABCD

Remeber if A is a friend to B, it implies B is also a friend to A.

It is given that There are exactly 4 people having 1 friend. So we need two pairs to satisfy this let it be A----B C----D

So it is clear that A ,B,C,D r the 4 pople having 1 friend only. Now we cannot select any one of these 4 for the other combinations(bcoz in that case they will have more than 1 friend)

It is given that 3 people have exactly 2 friends. So we need 3 pairs from remaing (E,F,G)

ie E-----F E----G F----G

Clearly E have only 2 friends. SimilarlyF and G also have only 2 friends.

So totally there are 5 pairs of friends.

Now 2 people from 7 can be selected in 7C2 ie 21 ways. Out of these 21 there r 5 pairs of friends. Excluding them we have another 16 pairs.