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 (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). 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

My answer is c - 4/7, correct ans is 16/21. pls tell me how???

The total no. of selection of 2 persons from 7 = 7c2 =21

lets number all the persons in the room as 1 2 3 4 5 6 7

possible friendship combinations is 1-5 , 2-5 , 3-6 , 4-7, 6-7 there will be 5 pair of friends which will satisfy the given condition so probability of selcting a pair of friends out of 21 pairs = 5/21

prob. of not selecting friends=16/21 :lol:

cityboy- thanks for the explanation. I understand how you got 21 total combinations, but i don't understand how you got the five different friend combinations. can you explain your logic? thanks very much.

Consider 1-2 a pair of friends, 3-4 another pair, and 5-6-7 the trio. This satisfies our conditions (4 people have 1 friend, 3 people have 2 friends).

You've already seen how to get to 21 possible combos. Cityboy finished the calculation by figuring out how many ways to pick friends and then subtracted from 21 to get how many ways to pick non-friends.

If we pick 1 first, then we have exactly one way to pick 1's friend: 2.
If we pick 3 first, then we have exactly one way to pick 1's friend: 4.
If we pick 5 first, then we have exactly two ways to pick 5's friend: 6 and 7.
If we pick 6 first, then we have exactly one way (that doesn't overlap with the above) to pick 6's friend: 7.

Those are the 5 ways to pick a friend, so there are 21-5 = 16 ways to pick non-friends. Make sense?

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

I knew this query was coming. Lets divide the 7 persons into 2 groups those who have 1 friends call this group A and those who have 2 friends call this group B .

LETS CALL THE MEMBER OF GROUP A 1 2 3 4 AND THOSE OF GROUP B 5 6 7
You can call them anything u like I am more comfortable dealing with numbers.

Now the only way in which the given condition that 4 persons have 1 friend and 3 have 2 friends can be satisfied is as follows

A B

1 5
2
3 6
4 7



Condition will be only satisfied when 2 members of group A have friendship with 1 member of group B say 1-5 & 2-5
Remaining 2 members of group A have bond with remaining 2 members of group B each say 3-6 & 4-7 and the last 2 members of group B i.e. 6 7 have bond with each other 6-7

In place of 1-5 & 2-5 u can have 3-5 & 4-5 or a variety of different other permutations but the guiding (bonding) principle will remain the same i.e. 2 memb. Of group A with 1 memb. Of group B and so on…

If u violate this rule then condition will not be satisfied .
U can have different permutations but in each scenario no. of friendship pair will be 5

SO out of 21 pairs 5 are friendship pair so the prob. Of not selecting friend =1- (5/21)=16/21

Hope it helps

You can also treat this more like a probability problem than a combinatorics problem. It might not be a consistent approach, but what the heck.

P(first person choosen has only one friend)*P(second person choosen is not a friend of first person, given that first person only has one friend)+P(first person choosen has two friends)*P(second person choosen is not a friend of first person, given that first person has two friends) = 4/7*5/6+3/7*4/6 = 20/42+12/42=32/42 = 16/21.

/Jeff

Hi everyone,

Which method of solving do you prefer or do you think is more reliable? I solved it Jeff's way, by using probabilities.

ManhattanGMAT experts - your thoughts?

Thanks,

Leon

the most important aspect of problem solving, especially with the draconian time restrictions imposed by the gmat, is FLEXIBILITY. it's not a very good idea to designate one approach or the other as 'more reliable', because, by so doing, you're essentially shutting the door on the other approach. the thing is, though, that it doesn't really matter which approach is more reliable, more elegant, or more anything on average. instead, what matters is the first legitimate approach you can get your hands on, whether that be combinatorial, probabilistic, or whatever else.
remember that everything on this exam is being done under severe time constraints, and act accordingly - try to learn ALL the possible approaches to problems, with as few prejudices as possible.

by the way, here's a way you can solve this problem without bothering to write out lists.

consider the problem from a basic probabilistic standpoint: you want the NUMBER OF PAIRS OF NON-FRIENDS, divided by the TOTAL NUMBER OF POSSIBLE PAIRS of people in the room.

to find the NUMBER OF PAIRS OF NON-FRIENDS (the numerator), you should find the number of pairs of friends first. to do this, you could make random lists, as is done in earlier posts in this thread. but here's another way:
just add up all the friendships that are listed. since four people have 1 friend in the room and three people have 2 friends in the room, that's a total of 4x1 + 2x3 = 10 friendships.
BUT
when you do this arithmetic, you're counting all the pairs of friends twice. (say A and B are friends. then you're counting that pair of friends once when you count A's friendships, and again when you count B's friendships. same goes for all the other pairs of friends.) therefore, the total number of different pairs of friends is actually half this number, i.e., 5 pairs of friends.
this is a nice alternative to listing out the friends - and it stays manageable, even if the problem features a much larger # of people (a situation that would make the listing method spiral out of control).

to find the TOTAL NUMBER OF POSSIBLE PAIRS of people (the denominator), take the # of different combinations of 2 people chosen from a set of 7. whether you use the anagram method (yynnnnn) or the combinations formula, this expression comes out as (7!)/(2!5!), or 21 total pairs.

finally, now that we have the total # of pairs, we can subtract to find the total # of pairs of non-friends: 21 - 5 = 16.

therefore, the probability is 16/21.

there

Let the 7 people be A, B, C, D, E, F, G.
Total No. of 2 way friendships: 7C2 = 21.

Arrange the people in the following way:
AB
BC
CD
DE

In above list, B, C, D have 2 friends each, so that satisfies the 2nd condition. A and E have just 1 friend each. Add FG to the list...Now A, E, F and G have one friend each satisfying the 1st condition.
So we have total no of friendships according to condition = 5
Total no. of 2 way non friends = 21-5= 16
Therefore, reqd prob is 16/21

This is the way I approached it. It just makes more sense to me and may be faster. However, this could be more error-prone as you really need to keep the numbers straight (i.e. remember to reduce the denominator after "picking out" a person and multiplying it with from the right "group"). Seperating out the groups takes some thinking, too, but that's common to both the probability and combinatorics apporach. Thanks Jeff!

Good work, all. Lots of ways to get to the answer!

I'm confused about this probability based approach, in the room we have 7 people, 2 who are friends with each other and 3 who are friends with each other, and 2 who are not friends with anyone.

Starting from here, how do we derive this probability? It seems like the (4/7)*(5/6) is based on the assumption that 4 persons out of the 7 have a friend each, which is not the case as there is double counting (friends with each other)?

Also why is the case where the first person chosen has no friends not one of the options? :?

Thank you!

-Praneeth

Praneeth, re-read the problem: "4 people have exactly 1 friend," so that's 2 pairs of people who are friends with each other, and zero people who are friendless. I think that is the main source of your confusion.

As for the "double counting" concern: This method has you picking individuals and then pairing them, not picking pairs. Some “double counting” on such pairing problems makes sense because there are two ways to pick A and B--you could pick AB or pick in the opposite order BA, doubling the chance of making that pair. To get your head around this, you might want to write out all 7*6=42 ways to select 2 people from 7 (that’s 21 possible pair selections, when you account for the double counting). Note that the denominators of 42 in the probability method account for this.

_________________
Emily Sledge
Instructor
ManhattanGMAT

I decided it would be easier to set the problem up this way:

probability that the second person is NOT the first person's friend = 1 - the probability that they are his friend

= 1 - [(prob. that 1st person has 1 friend)x(prob. that the second person is that friend) + (prob. that 1st person has 2 friends)x(prob that the second person is one of those two friends)].

= 1 - [(4/7)x(1/6) + (3/7)x(2/6)]

= 1 - [(4/42) + (6/42)]
= 1 - (5/21)
= 16/21

Alternatively, I could have just said (4/7)x(5/6) + (3/7)x(4/6) which is equal to 16/21 rather than using the whole 1 minus all the rest...