# combination

• Jan 27th 2008, 01:23 PM
shilz222
combination
$\displaystyle 8$ bottles of zinfandel

$\displaystyle 10$ bottles of merlot

$\displaystyle 12$ bottles of cabaret.

(a) If he wants to serve $\displaystyle 3$ bottles of zinfandel and serving order is important, how many ways are there to do this?

$\displaystyle \frac{8!}{5!} = 336$

(b) If $\displaystyle 6$ bottles of wine are to be randomly selected from the $\displaystyle 30$, how many ways are there to do this?

$\displaystyle \binom{30}{6} = 593,775$

(c) If $\displaystyle 6$ bottles are randomly chosen, how many ways are there two bottles of each variety being chosen?

(d) If $\displaystyle 6$ bottles are randomly chosen, what is the probability that this results in two bottles of each variety being chosen.

(e) If $\displaystyle 6$ bottles are randomly chosen, what is the probability that they are all the same variety?
• Jan 27th 2008, 01:36 PM
Jhevon
Quote:

Originally Posted by shilz222

(c) If $\displaystyle 6$ bottles are randomly chosen, how many ways are there two bottles of each variety being chosen?

$\displaystyle { 8 \choose 2} {10 \choose 2} {12 \choose 2}$

Quote:

(d) If $\displaystyle 6$ bottles are randomly chosen, what is the probability that this results in two bottles of each variety being chosen.
$\displaystyle \frac {{ 8 \choose 2} {10 \choose 2} {12 \choose 2}}{{ 30 \choose 6}}$

Quote:

(e) If $\displaystyle 6$ bottles are randomly chosen, what is the probability that they are all the same variety?
$\displaystyle \frac {{ 8 \choose 6} + {10 \choose 6} + {12 \choose 6}}{{30 \choose 6}}$

why is that?
• Jan 27th 2008, 01:42 PM
shilz222
There are $\displaystyle \binom{8}{6}$ ways to choose all zinfandel, etc..