Nikhil, simply take it this way.

nCr is a selection of n things from a total of r things.

For your problems,

1. Select 1 blue ball from 5 blue,4 Green & 3 Yellow Balls
This means. 1 blue ball out of 5 blue balls
and 1 ball out of total 12 balls.

Hence we write 5C1/12C1

2. Select 2 blue ball from 5 blue,4 Green & 3 Yellow Balls
This means. 2 blue ball out of 5 blue balls
and 2 balls out of total 12 balls.

So, we write 5C2/12C2 and on solving this is the same as what you have written.

What you are doing is:(5/12)*(4/11)

It simply means:
(5C1/12C1)*(4C1/11C1) = (1 blue balls from 5 blue balls and total 12 balls)*(1 blue ball from 4 blue balls and total 11 balls). Because the one ball has already been removed.

Does this help?

nikhilgmat31 wrote:Please suggest when to use nCr formula to select a item for a group.

lets says we need to select 1 blue ball from 5 blue,4 Green & 3 Yellow Balls. - then we can simple take 5/12
Then What is the need of writing 5C1/12 - however 5C1 also gives 5

In other case

lets says we need to select 2 blue ball from 5 blue,4 Green & 3 Yellow Balls. - then we can simple take 5/12 * 4/11 = 5/33

In that case do we need to write it as 5C2 = 10

What will be denominator in this case-- is it 12

so probability is 5C2/12 = 10/12 = 5/6

Please suggest . I am confused.

You are messing up combination and probability together. Both these concepts are used together is solving probability questions.

Post questions related to combination and probability ... that would be a good point to start.

solving question is another part of game. I want to be rich in concepts of counting, probability, combinations & permutations.

I am mixing all these 4.

OptimusPrep wrote:Nikhil, simply take it this way.

nCr is a selection of n things from a total of r things.

For your problems,

1. Select 1 blue ball from 5 blue,4 Green & 3 Yellow Balls
This means. 1 blue ball out of 5 blue balls
and 1 ball out of total 12 balls.

Hence we write 5C1/12C1

2. Select 2 blue ball from 5 blue,4 Green & 3 Yellow Balls
This means. 2 blue ball out of 5 blue balls
and 2 balls out of total 12 balls.

So, we write 5C2/12C2 and on solving this is the same as what you have written.

What you are doing is:(5/12)*(4/11)

It simply means:
(5C1/12C1)*(4C1/11C1) = (1 blue balls from 5 blue balls and total 12 balls)*(1 blue ball from 4 blue balls and total 11 balls). Because the one ball has already been removed.

Does this help?

Awesome explanation Ankur. Thanks so much.

Can you please provide some reference to guide to counting(combinations & permutations) & probability


Thanks in Advance.