# C2 Ways in Urn Problem ?

• May 8th 2010, 04:53 PM
mathnoob123
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
$\displaystyle nC_k$ or $\displaystyle {n\choose k} = \frac{n!}{(n-k)!k!}$

This means combination. If we have a total of $\displaystyle n$ elements and we want to select a group of $\displaystyle k$ of them. Then there are $\displaystyle nC_k$ 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:
$\displaystyle 5!=5*4*3*2*1=120$

Anonymous1 already explained the combination part.

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

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

$\displaystyle {12\choose 2} = \frac{12!}{(12-2)!2!}= \frac{12!}{10!*2!}=66$

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 :)