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In a room filled with 7 people, 4 people have exactly 1

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In a room filled with 7 people, 4 people have exactly 1 [#permalink] New post 27 Sep 2006, 08:31
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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 [#permalink] New post 27 Sep 2006, 09:59
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.
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 [#permalink] New post 27 Sep 2006, 10:27
Futuristic ,
could you pls care to walk us thru the solution ?
i am still not clear :(
thanks
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 [#permalink] New post 27 Sep 2006, 15:40
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.
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 [#permalink] New post 27 Sep 2006, 16:10
I think the answer is 16/21, if its correct, I can try to explain....
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 [#permalink] New post 27 Sep 2006, 17:11
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
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 [#permalink] New post 27 Sep 2006, 17:13
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
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 [#permalink] New post 27 Sep 2006, 17:46
Consider:

a-b
b-c
c-d
e-f
f-g

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
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 [#permalink] New post 27 Sep 2006, 18:03
this makes sense, I think I over complicated my method just to get 16/21

thanks futuristic
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 [#permalink] New post 27 Sep 2006, 18:07
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.
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 [#permalink] New post 27 Sep 2006, 18:17
you are right captain, I was wrong
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 [#permalink] New post 27 Sep 2006, 23:39
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
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.

So prob = 16/21

I think i am clear
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Last edited by cicerone on 25 Sep 2008, 00:17, edited 1 time in total.
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 [#permalink] New post 28 Sep 2006, 09:06
Thanks Cicerone for the explanation.
  [#permalink] 28 Sep 2006, 09:06
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