# Doubt in permutation/combination from AV casino

#1

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
so,
12C4 * 8C4 * 1

Why would you divide it by 3!

#2

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.

#3

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

#4

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…

12c4,8c4,4c4
other 4 also can be created by 12c4…

Chk tis out: