Doubt in permutation/combination from AV casino


Hi All,
Sorry for so silly doubt , but didn’t understand solution of below problem.

Q. How many ways are there to split a dozen people into 3 teams, where all three teams have 4 people each?

As per my understanding , 1st team can be selected in 12C4 ways.
2nd team in 8C4 and 3rd in 4C4
12C4 * 8C4 * 1

Why would you divide it by 3!

Please expain?


Another doubt…, How to solve below question?

Question :
After a lot of tussle the couples were broken into 3 teams with 2 couples each i.e. Team A has couple 1 and couple 2, Team B has couple 3 and couple 4 and Team C has couple 5 and couple 6. Each person throws a 6 sided fair die. The couple with the rarest sum wins. What is the probability of obtaining the rarest sum?
For example: Say Couple 1 throws 1 & 2, then their sum is 1 + 2 =3. The number of possible ways 3 can be obtained is (1,2) or (2,1) so two ways. Say Couple 2 throws 1 & 1, then their sum is 2. The number of possible ways 2 can be obtained is (1,1) which is just one way. So 2 is a rarer sum than 3. You need to find what the rarest sum is first and then obtain it’s probability.


First you have to find the rarest sums in two dice rolling i.e from 36 sample spaces.

so , it is (1,1) and (6,6)… 2 chances out of 36… ans : 1/18


That is about fixing the groups in a particular ORDER.

If order is not there, you can keep your groups in any ways you want…

in this case, there are 6 ways you can create ur groups…

other 4 also can be created by 12c4…

Chk tis out: