1. ## Permutation trouble

Hello,

I am wondering whether I solved the following correctly.

"How many license plates containing seven characters can be created from letters and numbers if the first two characters must be uppercase letters and the last five characters must be digits from 0 to 9? Repetition of numbers is allowed. Repetition of letters is not allowed".

Here's my solution.

1) I assumed a universal set for the license plate as U = {uppercase letter, uppercase letter, number, number, number, number, number}.
2) For the first two characters I assumed a sample set A ={A,B,C...Z}, for the last five characters there's the sample set B = {0,1,2,3...9}.
3) Considering the repetition of letters is not allowed and that the order matters in set U, I determined 650 possible permutations in set A, following the formula n!/(n-r)!, which comes down to 26!/(26-2)! = 650 permutations. So set A has 650 possible permutations.
4) Considering the repetitions of numbers is allowed and that the order in set U matters, I determined 100.000 possible permutations in set B, following the formula n^r, which comes down to 10^5 = 100.000 permutations. So set B has 100.000 possible permutations.
5) Now set A and B are added together, giving what I think are 100.650 possible license plates.

Are my thoughts correct or did I go wrong somewhere? It feels like I went wrong somewhere. Thanks in advance =)

2. ## Re: Permutation trouble

Originally Posted by Wolfy1984
"How many license plates containing seven characters can be created from letters and numbers if the first two characters must be uppercase letters and the last five characters must be digits from 0 to 9? Repetition of numbers is allowed. Repetition of letters is not allowed".

Here's my solution.

1) I assumed a universal set for the license plate as U = {uppercase letter, uppercase letter, number, number, number, number, number}.
2) For the first two characters I assumed a sample set A ={A,B,C...Z}, for the last five characters there's the sample set B = {0,1,2,3...9}.
3) Considering the repetition of letters is not allowed and that the order matters in set U, I determined 650 possible permutations in set A, following the formula n!/(n-r)!, which comes down to 26!/(26-2)! = 650 permutations. So set A has 650 possible permutations.
4) Considering the repetitions of numbers is allowed and that the order in set U matters, I determined 100.000 possible permutations in set B, following the formula n^r, which comes down to 10^5 = 100.000 permutations. So set B has 100.000 possible permutations.
5) Now set A and B are added together, giving what I think are 100.650 possible license plates.
The results are multiplied not added: $(650)\cdot(10^5)$

3. ## Re: Permutation trouble

For the first letter you have 26 possible choices. For the 2nd letter you have 25.

For each of the numbers you have 10 choices. There are 5 of these.

so the total number of unique license plates is given by

$N = 26 \cdot 25 \cdot 10^5 = 65,000,000$