4 monitors each day

This topic has expert replies
User avatar
Master | Next Rank: 500 Posts
Posts: 142
Joined: Mon Jan 10, 2011 8:03 am
Thanked: 19 times

4 monitors each day

by krishnasty » Sun Jun 19, 2011 5:27 am
In a class in a certain school, there are 25 students. Four students are appointed monitors every day. At the end of a term, it is found that every possible pair of students have been monitors together exactly once. How many days are there in the term?
OPTIONS

1) 30
2) 50
3) 60
4) 75
---------------------------------------
Appreciation in thanks please!!
Source: — Problem Solving |

Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Sun Jun 19, 2011 6:30 am
Hi,
This is a tough one.I am not sure whether my reasoning is correct. I will make an attempt to solve this.
4 monitors can be selected in 25C4 ways
In any combination of 4 if we select a pair, the other two can be selected in 23C2 ways.
i.e, each pair is repeated 23C2 times.
But, here each pair should be present only once.
So, one out of every 23C2 pairs is favorable.
So, the number of days = 25C4/23C2 = 50

(or)
Number of pairs that can be selected is 25C2 = 300
Consider a combination 1234
It can be formed from
12, 34
13, 24
14, 23
So, any combination of 4 elements can be formed from 6 pairs. So, 6 pairs form the same combination.
So, number of days = 300/6 = 50.


Hence, 2

Can you post the source,OA and OE?
Cheers!

Things are not what they appear to be... nor are they otherwise

User avatar
Master | Next Rank: 500 Posts
Posts: 142
Joined: Mon Jan 10, 2011 8:03 am
Thanked: 19 times

by krishnasty » Sun Jun 19, 2011 7:27 am
Frankenstein wrote:Hi,
This is a tough one.I am not sure whether my reasoning is correct. I will make an attempt to solve this.
4 monitors can be selected in 25C4 ways
In any combination of 4 if we select a pair, the other two can be selected in 23C2 ways.
i.e, each pair is repeated 23C2 times.
But, here each pair should be present only once.
So, one out of every 23C2 pairs is favorable.
So, the number of days = 25C4/23C2 = 50

(or)
Number of pairs that can be selected is 25C2 = 300
Consider a combination 1234
It can be formed from
12, 34
13, 24
14, 23
So, any combination of 4 elements can be formed from 6 pairs. So, 6 pairs form the same combination.
So, number of days = 300/6 = 50.


Hence, 2

Can you post the source,OA and OE?
OA is indeed 2..
I have a collection of GMAT material given to me by a friend. just OA is given but no OE.
i still didnt understand the solution.can experts help?
---------------------------------------
Appreciation in thanks please!!