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 Posted: Sat Apr 21, 2007 7:05 pm
Okay so my class has just finished learning about factorials and combinations and the P thing. I forgot the name.
Okay, the problem is this; I have a test and I mostly slept through the whole combinations thing and we can't use calculators, so I was wondering if someone could simply show me how to do problems such as this:
60C3
15C3
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Posted: Sun Apr 22, 2007 11:20 am
Permutation:
nPa =
n!
(n-a)!
Combination:
nCa =
n!
a!(n-a)!
Factorial:
n! = n x (n-1)!
(or n = n x (n-1) x (n-2) x ... x 1)
For these problems, it's often useful to write n! = n x (n-1) x (n-2) x ... x (n-a)!, as you'll see.
So, for your specific problem, we start by plugging each into the formula:
60!
(3!)(57!)
_________
15!
(3!)(12!)
Get rid of that nasty compound fraction by multiplying by
(3!)(12!)
15!
______________
(3!)(12!)
15!
which cancels out the bottom.
This gives us
(60!)(3!)(12!)
(3!)(57!)(15!)
Ok, the 3! cancels, so we can get rid of that. Let's rewrite 60! as (60 x 59 x 58 x 57!), and 15! as (15 x 14 x 13 x 12!)
(60 x 59 x 58 x 57!) x 12!
57! x (15 x 14 x 13 x 12!)
Now we can drop the parenthesis and cancel the 57! and the 12!:
60 x 59 x 58
15 x 14 x 13
Simplify...
4 x 59 x 29
7 x 13
Basically, all you need to do for these problems is convert nPa or nCa to factorials, rewrite the factorials so stuff cancels, and simplify.
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Posted: Sun Apr 22, 2007 3:54 pm
Ooh, I see what I've been doing wrong. Thanks.
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