There are n new students in a class. Among any three of them, there ex

HoneyLemon

I am sure you will never get such a question in the actuals, but it is always good to understand different concepts. Not an easy one by any standard.
But all the above solutions are way off the mark.

Two conditions
1) Pick any 3 of them and you will have two who know each other.
So there will be groups of people who know one another within the group. Because if you have three isolated groups, you can choose one from each group and none of three would know each other.
2) Pick any 4 of them and you will have two who do not know each other.
This restricts the number of people in any one group to 3
Because if there are 4 in the group, you can pick up these 4 and all would know each other.

With these two conditions, the immediate answer could be 6 : 3 in each group
ABC and DEF

But we are looking for max number, so we can think of interconnected groups.
Say, in each new group, two remain the same and one is changed. But we cannot have more than 2 completely isolated groups.

So groups can be
ABC
BCD
CDE
DEF
EFG
FGH
GHA

Now, we don’t go for next group GHI, that is we do not introduce new member I.
Because then we will have 3 isolated groups : ABC, DEF and GHI. The moment we pick up one from each of these 3 groups, we will not have two people known to each other. Example ADG or AEH and so on.

Now, we can get back to our two conditions and see if they are fulfilled.
(1) Pick any three : Only two groups will be completely isolated. So the third student picked up will be related to at least one of these two groups.
(2) Pick any four : As we have groups of only 3 people who are known to one another, the fourth person added will not know at least one of them. Example ABCD - A and D are not known to each other. OR ABCF - Again F does not know A, B or C.

Total students - A, B, C, D, E, F, G and H, so 8 students. You can also draw a circle and solve, but this is much easier to grasp this way.

B
_________________

giving a try

Among any three of them, there exist two who know each other ; 3 and among any four of them, there exist two who do not know each other
3c2*2+2c1 = 8
IMO B

Hi Archit3110

Please elaborate your answer.
Please provide similar questions and methods to solve them.

Hi Bunuel
Please provide OE for the above question.

Can someone help with an answer?

For me is really difficult, I do not know were to start

Thanks in advance,
Juan

[quote="Bunuel"]There are n new students in a class. Among any three of them, there exist two who know each other and among any four of them, there exist two who do not know each other. Find the greatest possible value of n.

A. 7
B. 8
C. 9
D. 10
E. 11

jcgomezlv Kinshook

To my understanding
1) Among any three people two know each other
Lets assume three people as X X Y - 2 people know each other
2) Among any four people two do not know each other
Assuming four people as X X Y Z - Author mentions about Y and Z

Now combining together the rule to get maximum number of students
We might mistakenly assume
XXY & XXYZ and come up with 7

But here's the catch the class can have
2 sets of XXYZ because from this set if you pick 3 you can arrive at XXY (XXY is a subset of XXYZ)
so the maximum possible class value is XXYY XXYY -> 8 members

Bunuel please explain or post actual solving method.

chetan2u May you please help on this .

chetan2u Indeed a complex one .. Thank you for such a comprehensive explanation .

Can't help but believe there is fuzzy language in the question.

As it reads, "only" two people know each other in any group of three.

If I assign A,B and C as a group of 3 and declare that A and B know each other, then neither A nor B know C, two groups of two not knowing each other.

Once we add a fourth person D, the above immediately conflicts with the statement that one group of two doesn't know each other in any group of 4.

Posted from my mobile device

Here's how I approached this question.

2 students know each other in a group of 3, and 2 do not know each other in a group of 4.

So, in a group of, AABB is possible, however, AABC is not possible. Therefore all students operate in groups of 2s. Hence the answer should be even and should be a multiple of 4.

Hence, 8 is the correct answer.

This might not work if the options change.

Unfortunately, the question stem states "in any group of 4...", which suggests AABC should be a possibility. But as you point out, it isn't.

So the language of the question has internal conflicts.

Posted from my mobile device