Question

My first probability question :P
In how many ways can you choose a group of three people from a group of six?

Answer 1

(will type attempt in a bit)

Answer 2

\[\frac{n!}{r!(n-r)!}\]
so... seeing that this is a combination type problem, I have to use that formula
if I let n = 6 and r = 3, then

\[\frac{6!}{3!(6-3)!}\]
\[\frac{6!}{3!(3)!}\]
\[\frac{720}{36} = 20\]
so there's 20 different ways to choose a group of 3 from 6.

Answer 3

I think that's right???? I mean the combination formula is straight forward.

Answer 4

@agent0smith geez man...how long has it been since I saw you on OpenStudy?

Answer 5

can somebody help me on mines

Answer 6

oh yeah like posting your question on mine will help smh.

Answer 7

that seems right
"In how many ways can you choose a group of three people from a group of six?
"
There is 6 ways to choose the first
There is 5 ways to choose the second
There are 4 ways to choose the third
and then we need to figure out how we can arrange the 3 objects so the above product over 3!

\[\frac{6 \cdot 5 \cdot 4}{3!} \text{ which is the same as } \frac{6 \cdot 5 \cdot 4 \cdot 3!}{3! \cdot 3!} =\frac{6!}{3! \cdot 3!}\]

Answer 8

ah thanks. I haven't touch probability in forever which is why I'm shaky at it.

Answer 9

I can understand why you would be shaky. It is hard to count especially when the numbers get really big.

Answer 10

And the orders get crazy.

Answer 11

yeah... but I haven't even touch this subject at all. xD! Apparently Discrete Math has this... >_<

Answer 12

so it will be like
6! = 6 x 5 x 4 x 3 x 2 x 1
over 3! = 3 x 2 x 1

and at least the 3 x 2 x 1 cancels
6 x 5 x 4 / 3 x 2 x 1 = 120/6=20.. ah I see ...