# Math Help - combinations vs permutations questions

1. ## combinations vs permutations questions

hi, i'm new here and am looking for some help!

this was in a chapter that uses permutations and combinations...but i cant figure out how to get it. it's just for practice, and i know the answers (the probability of 5 out of 52 is higher)...but i'd like to find out why.

Are the odds of winning in a lotto drawing higher when picking 6 numbers out of 49 vs. 5 numbers out of 52? calculate each probability.

This one's more wordy but I'm also confused:

In June 2004, Canada introduced a change in its lottery that violated the usual convention that the smaller the probability of an event, the bigger the prize. In this lottery, participants have to guess six numbers from 1 to 49. Six numbers between 1 and 49 are then drawn at random, plus a seventh "bonus number."

a) a fifth prize of $10 goes to those who correctly guess exactly three of the six numbers, but do not guess the bonus number. Find the probability of winning fifth prize. b) A sixth prize of$5 goes to those who correctly guess exactly 2 of the six numbers plus the bonus number. Find the probability of winning sixth prize, and compare this with the probability of winning fifth prize.

Again, I know the answer for a) is .01642 and b) is .01231...but if someone could show me HOW to get to it that'd be great!

Thanks! This seems like a great community

2. First share what you have tried to do in order to get to the solution so someone could tell you what went wrong.

3. sure, i'll be happy to share

for the first one, i have 49 over 6 (the combination form) divided by 49! and 52 over 5 divided by 52!

I know it's probably not right, but i'm not really sure where to even begin! same with the second. what should i do? thanks!!

4. Originally Posted by plan21
sure, i'll be happy to share

for the first one, i have 49 over 6 (the combination form) divided by 49! and 52 over 5 divided by 52!

I know it's probably not right, but i'm not really sure where to even begin! same with the second. what should i do? thanks!!
Some better ways to write "49 over 6 (the combination form)" in text would be

49 choose 6
C(49,6)
49C6

Also we have a subforum for LaTeX Help to get nice typesetting like

$\displaystyle \binom{49}{6}$

So for the first, your calculations unfortunately do not make sense. First scenario we must guess one combination out of C(49,6) possible combinations, so the probability is 1 / C(49,6). Compare this with 1 / C(52,5).

For the second, try to reason it out. For example in (a), how many ways are there to guess exactly three correct numbers out of 6? And after fixing those first 3, how many ways are there to select the remaining 3?