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Intern
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In a room filled with 7 people, 4 people have exactly 1 [#permalink]
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27 Sep 2006, 09:31
This topic is locked. If you want to discuss this question please repost 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?
5/21
3/7
4/7
5/7
16/21
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Director
Joined: 28 Dec 2005
Posts: 752

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.



Intern
Joined: 27 Aug 2006
Posts: 32

Futuristic ,
could you pls care to walk us thru the solution ?
i am still not clear
thanks



Manager
Joined: 15 Aug 2006
Posts: 54

Ans. D
The question can be pictured in the following manner where A,B,C...G denotes each of the individuals.
A>D (A is friend of D)..
A>E
A>F
A>G
A>B>C (A is a friend of B and B is a friend of C).
A>C>B
Probability of picking 2 individuals from 7 who are not friends =
favourable/Total
Favourble individuals who when picked will not be friends are D,E,F,G and (B or C) = 5 individuals
Total number of ppl = 7
so Prob = 5/7.
Hope I m correct.



Director
Joined: 28 Dec 2005
Posts: 752

I think the answer is 16/21, if its correct, I can try to explain....



Manager
Joined: 28 Aug 2006
Posts: 243
Location: Albuquerque, NM

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



Manager
Joined: 28 Aug 2006
Posts: 243
Location: Albuquerque, NM

God, hope I don;t get such brain teasers in real exam, they take a long time, unless you luckily get the right idea at the start



Director
Joined: 28 Dec 2005
Posts: 752

Consider:
ab
bc
cd
ef
fg
From the above:
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
total = 16
total without restrictions = 7c2 = 21
Prob = 16/21



Manager
Joined: 28 Aug 2006
Posts: 243
Location: Albuquerque, NM

this makes sense, I think I over complicated my method just to get 16/21
thanks futuristic



Intern
Joined: 02 Aug 2006
Posts: 23
Location: NYC

jainan24 wrote: 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
Thus 5+5+3+3 = 16/21.



Manager
Joined: 28 Aug 2006
Posts: 243
Location: Albuquerque, NM

you are right captain, I was wrong



Senior Manager
Joined: 28 Aug 2006
Posts: 304

Folks, let's keep this simple. A,B,C,D,E,F,G are the seven friends. 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 AB CD 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 EF EG FG 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. So prob = 16/21 I think i am clear
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Last edited by cicerone on 25 Sep 2008, 01:17, edited 1 time in total.



Manager
Joined: 15 Aug 2006
Posts: 54

Thanks Cicerone for the explanation.










