Thread: C2 Ways in Urn Problem ?

1. 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

2. $\displaystyle nC_k$ 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?

3. 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

4. 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

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

Cheers Guys