A B C D E F
There can be only 15 unique pairs from 6 different people and the order does not matter in the pair since its like picking a commitee.

3 pairs can be picked from 15 pairs in 15c3 ways.

Then these 3 pairs can be arranged in 3! i.e 6 ways in the continents

6*15c3 = 2730


Stuart/Ian or others please feel free to share your thoughts.


My question is if:

AB is assigned to continent 1
CD to continent 2
EF to continent 3

AND

AB is assigned to continent 1
DF assigned to continent 2
CE assigned to continent 3

considered as 2 different ways. I think so and hence my 2730 answer. Please correct me if I am mistaken.

Firstly,

Dont have an OA, sorry.

Secondly,

AB is assigned to continent 1
BD assigned to continent 2
CE assigned to continent 3

I have a question. If B is in continent 1 how can B be in continent 2 at the same time. So, according to me, we cannot have a combination of AB and BD at the same time.

Thirdly, my solution

A B C D E F

Number of groups that can be formed from 6 people: 6C2 = 15
Number of ways of assigning 15 groups.

Continent 1: We have 15 ways of selecting 1 group for this continent (15C1)

Continent 2: For instance, if we selected pair AB for Cont 1, we can choose our next group from the remaining 4 people (C D E F). So number of ways of selecting 2 people from 4 = 4C2 = 6

Continent 3: For this continent we'll have only 2 people remaining so just one group.

So total number of ways of arranging 15 groups into 3 continents = 15*6*1 = 90

Just my opinion.

I am not good with Permutations, Combinations & Probability. But, will give my shot at this question

3 pairs can be formed using 6 students in 20 ways (6c3). The 3 pairs can be assigned to 3 different continents in 6 ways (3!). So, adding 20+6=26ways is my answer.

I hope I am correct

Thanks Mals and Vemuri for attempting it.

I have a question. If B is in continent 1 how can B be in continent 2 at the same time. So, according to me, we cannot have a combination of AB and BD at the same time.

Mals, sorry that was a typo and have corrected it.

Mals, I think u may be right. Good solution.

Hoping Ian/Stuart can confirm this also.

Regards,
Cramya

Okay, let's learn some concepts by playing with them:

WHAT IF,

if we decide that continents are not moving anywhere

1 2 3

and we will assign people to them:

AB CD EF

so they can be listed as 3!= 6 ways

how many different pairs can we get ? 6C2= 15

so 15x6=90

Do you agree with this logic ? If not, why ?

Thanks Ian,Logitech, Mals and Vemuri.