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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 
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Stanley 19 Feb 2015 2:53 AM
Thanks a lot Guys!
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nandhinirajan 04 Feb 2015 12:56 PM
it`s very nice and i`m clear now.
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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? 
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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
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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...


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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
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sayali 12 Nov 2014 5:58 AM
thanks .....a very beautiful explanation

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Mohamed Suhail 06 Nov 2014 8:54 PM
this is perfect!! thanks a million :D
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nadan naik 30 Sep 2014 6:11 PM
nice super explanation
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