# Thread: Quick Question one Permutations/Combinations

1. ## Quick Question one Permutations/Combinations

Hey all, I had a few questions regarding some probability theory which have me completely stumped, would someone mind pointing me in the right direction?

Question #1-

Assume that all license plates in Kentucky consist of three numbers followed by three letters.
How many ways are there to get a Kentucky license plate with the number 41 appearing (without any digits between) and with the first two letters of your name appearing consecutively in order?

(If the first two letters of your name are the same you should calculate this problem using any two distinct letters.)

Question #2 -

Suppose a conference has 12 basketball teams. At the end of the season the four best teams play for the championship. How many different sets of four teams could be in the championship?

2. ## Re: Quick Question one Permutations/Combinations

Originally Posted by Danny09
Hey all, I had a few questions regarding some probability theory and have found myself completely stumped, would someone mind pointing me in the right direction?

Question #1-

Assume that all license plates in Texas consist of three numbers followed by three letters.
How many ways are there to get a Kentucky license plate with the number 41 appearing (without any digits between) and with the first two letters of your name appearing consecutively in order?

(If the first two letters of your name are the same you should calculate this problem using any two distinct letters.)

Question #2 -

Suppose a conference has 12 basketball teams. At the end of the season the four best teams play for the championship. How many different sets of four teams could be in the championship?
By the way the first problem is missing some information. (by the way when did we leave Texas and get to Kentucky?) :P

Now onto more serious matters. For the 41 part we can look at the problem like this. If we just had to count the number of ways to arrange 3 digits we would have _ _ _

$\displaystyle 10 \cdot 10 \cdot 10=1000$ ways to do it.

But now we want to fix two digits in order to get 41 so we have

41 _ or
_ 41

where we can put 0-9 in the _ this can happend a total of 18 ways. Now use the fundamental counting principle to finish part 1. Use the same trick for 2nd part.

What have you tried for the 2nd problem. Where are you stuck?

3. ## Re: Quick Question one Permutations/Combinations

Apologies for the delayed reply, thanks very much for the help on that first problem, I think I've got it now!

As far as the second problem, I know it requires a permutation, but unsure if I'm using the correct formula: n!/(n-k)! = n X (n-1) X ... X (n-k+1) where n = the total amount of observations and k = the chosen items.