Six three-representative delegations attend an international conferenc

Thanks Ruchi for writing this "There are 6 delegations, each contains 3 representatives." Now I can comprehend the ques.
(5*3*3*6)/2=135

I think answer should be D.
Total number of members: 6 * 3 = 18
Total number of handshakes(including among same delegations) : 18C2 = 153
Now, Total handshakes among same delegations = 6 * 3C2 = 18
So, Answer = 153 - 18 = 135

There is another way to solve this: each member shakes hand with other members and not with herself, therefore each member shakes hands with 15 members. So, 18 members shake hands with 15 others= 18*15. But when one member shakes hand with other, the other will not be required to repeat the same (since, shaking hands is a two way thing, does not need to be repeated once done between two members. Hence, 18*15/2= 135.

We are given that there are 3 representatives from 6 different delegations. So, there are a total of 18 representatives.

If every representative were to shake hands with all other representatives (meaning all 18 reps would shake hands), this would happen in the following number of ways:

18C2 = (18 x 17)/2! = = 9 x 17 = 153 ways

However, since each person shook hands with every person not from his or her own delegation, we can subtract out the number of times those handshakes occurred.

Since each company has 3 reps, the number ways those three reps can shake hand is 3C2 = (3 x 2)/2! = 3 ways, and since there are 6 companies, this would occur 6 x 3 = 18 times.

Thus, the number of ways for the reps to shake hands with every person not from his or her own delegation is 153 - 18 = 135 ways.

Alternate Solution:

Let’s count the total number of handshakes. Since each of the 6 x 3 = 18 representatives shake hands with 5 x 3 = 15 other representatives, this makes 18 x 15 = 270 handshakes. However, we counted each handshake twice, once for each of the parties involved in the handshake. Therefore, the number of handshakes that take place is 270 / 2 = 135.

Answer: D

total handshakes would 18c3 ; 153
total hand shakes if done within each group ; 3c2 *6 ; 18
so total handshakes shakes hands only once with every other attendant except with those of his/her delegation ; 153-18 ; 135
IMO D

If every single person took turns shaking hands with all other people, we would have 18 * 15 handshakes in total (15 because "each delegate shakes hands only once with every other attendant except for those of his/her delegation"). However, with this product we would have any two people (of different delegation) shaking hands with each other twice, e.g. A shakes hands with B and B shakes hands with A are both counted with this method. We only need one of each combination to happen therefore we divide by 2 to get 9*15 = 135.

Ans: D

Given: Six three-representative delegations attend an international conference. The representatives shake hands when they are introduced to one another.
Asked: How many handshakes are possible if each delegate shakes hands only once with every other attendant except with those of his/her delegation?

Every delegate will shake hands with = 5*3 = 15 delegates
Since there are 18 delegates, total handshakes = 18*15 = 135*2

Since each handshake is counted twice, total handshakes = 135

IMO D

1. Total Handshakes possible = N(N-1)/2 ==> (18*17)/2 ==> 153

2. In one group there can be 2 hand shakes from one person to other 2 people say delegation one has ABC as members
a handshakes with AB,AC so other 2 also handshake with others with in delegation

3. total 6 hand shakes with in delegation divided by 2 will eliminate counting twice
hence==> 3 hand shakes per delegation * 6 delegations = 18 hand shakes which should not be counted

4. Total -handshakes with in delegations ==>153-18==> 135

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