C2 Ways in Urn Problem ?

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May 8th 2010, 04:53 PM
mathnoob123

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C2 Ways in Urn Problem ?

Hey Guys, i juz need some explanation about this problem which i attached as an image file ... as you can see the problem is already solved, i'm only concerned with the explanantion ...

1) what is meant by 12C2 & 8C2 ways ?
2)can u plz explain the terms that i highlighted with red ink like 8x7x6/2x1x6 and also 28/66 and etc.,

Thanks :)
May 8th 2010, 09:45 PM
Anonymous1

$nC_k$ or ${n\choose k} = \frac{n!}{(n-k)!k!}$

This means combination. If we have a total of $n$ elements and we want to select a group of $k$ of them. Then there are $nC_k$ or "n choose k" ways to combine them.

What exactly do you want explained about the highlighted problems?
May 9th 2010, 07:32 PM
mathnoob123

thanks for taking time and explaining the formula, "can u plz explain all the steps of this problem" ...

1)how did this step happened ..." 8! / (2!x6!) = 8x7x6! / 2x1x6! ".
2)how did i got 66 in the denominator of "28/66"
3)why there's an exclamation point (!) ...???

Thank You :)
May 9th 2010, 10:35 PM
downthesun01

3)The exclamation point represents a factorial. For example:
$5!=5*4*3*2*1=120$

Anonymous1 already explained the combination part.

1)All they did was expand $8!$ and the $2!$
$8!=8*7-6*5*4*3*2*1$ and $2!=2*1$

2)66 is the number of ways that two balls can be chosen from the urn. It's found by:

$<br /> <br /> {12\choose 2} = \frac{12!}{(12-2)!2!}= \frac{12!}{10!*2!}=66<br />$

This is explained in the solution to the problem that you posted
May 10th 2010, 01:35 AM
mathnoob123

Thank you very much downthesun01, Great Explanation ... i'm all cleared out with my doubts now ... and thanks to Anonymous1 too ...

Cheers Guys :)

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