Originally Posted by

**mrsenim** Hello,

I am new to probability.

Please have a look at the following question.

We have n = 10 persons and we wish to divide them at random into 3 groups consisting of 5, 3, and 2 persons respectively. In how many ways is this possible?

I have tried to solve it in following way.

Total number of possible ways =

$\displaystyle

10C5 = \left( 10! \over 5!(10-5)! \right) = \left( 10 \times 9 \times 8 \times 7 \times 6 \over 5 \times 4 \times 3 \times 2 \right) = \left( 30240 \over 120 \right) = 252

$

Next!

$\displaystyle

5C3 = \left( 5! \over 3!(5-3)! \right) = \left( 5 \times 4 \over 2 \right) = \left( 20 \over 2 \right) = 10

$

Next!

$\displaystyle

2C2 = \left( 2! \over 2!(2-2)! \right) = 1

$

$\displaystyle

S = 10C5 \times 5C3 \times 2C2 = 252 \times 10 \times 1 = 2520

$

I think the above solution is correct but I really don't know how it happens.

I mean to say that I am confused in

1 - Why are we multiplying 10C5 with 5C3 and 2C2 ? Which rule or formula implies this multiplication of combinations?

2 - When and in which situations do I need to multiply combinations?

Any help would be highly appreciated.