### Basics Concepts and Formulas in Permutations and Combinations

Comments(24)

Ilham
27 Mar 2015 10:30 PM

How nice; you lead the thoughts in step by step mode

nandhinirajan
04 Feb 2015 12:56 PM

it`s very nice and i`m clear now.

Shiksha
13 Jan 2015 3:47 AM

Hi,

I have a question, if we have 6 white balls , 3 blue balls and 2 Red balls and we have to pick a selection 4?

a) how many combinations it will be?

b) if always 2 white have to be selected from 4 then how many combinations possible?

Gaurav Dobriyal
24 Jan 2015 1:10 AM

6 3 2

w b r

2 1 1 = No. of ways arrangement is done.

1 2 1

1 1 2

Thus 6c2+3c1+2c1+6c1+3c2+2c1+6c1+3c1+2c2 = 15+3+2+6+3+2+6+3+1=41

if always 2 white then

6c2+3c1+2c1 = 15+3+2=20

Rajesh
25 Jan 2015 2:15 PM

But shouldn't it be 11c4 for (a).

the combinations could also be

6 3 2

w b r

0 3 1

4 0 0

2 2 0

and so on.. Nowehere it is mentioned that all colours need to be picked... So it can be done by 11C4 rite..?11c4=330 ..

and for (b) shouldn't it be 6c2*3c1*2c1 + 6c2* 3c2 + 6c2*2c2

Kindly correct me if I am wrong...

Gaurav Dobriyal
25 Jan 2015 2:48 AM

sorry the signs went wrong this one is correct

6c2*3c1*2c1+6c1*3c2*2c1+6c1*3c1*2c2 = 90+36+18 = 144

and

if always 2 white then

6c2*3c1*2c1=15*3*2=90

sayali
12 Nov 2014 5:58 AM

thanks .....a very beautiful explanation

Mohamed Suhail
06 Nov 2014 8:54 PM

this is perfect!! thanks a million :D

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