Three different letters are selected from a group of 26, order being important - this is a permutation of three elements out of twenty-six. These permutations number .

There are then four numerals, each chosen from a set of ten. Since repeats are allowed, the number of ways this can be done is 

By the multiplication principle, there are 

possible license plates.

There are 24 ways to choose an allowed letter (since two of the 26 are excluded) and 10 ways to choose an allowed digit. Since repeats are allowed, and four letters and three digits will be chosen, there will be

possible license plate numbers.

Each of the four letters can be selected twenty-four ways, since two letters are forbidden and since repeats are allowed. Similarly, each of the three digits can be selected eight ways for the same reason.

By the multiplication principle, the number of possible license plate numbers is

Each character of the licence plate number can be chosen from a set of 36 (26 letters plus 10 digits), and any character can be chosen multiple times. By the multiplication principle, this would allow  different licence numbers. But licence numbers that comprise only letters and those that comprise only digits are disallowed, so by similar math, this reduces the possibilities by  and , respectively. 

Therefore, there are  allowed licence plate numbers.

She has 4 choices for the shirt, 2 choices for the skirt per shirt chosen, and 3 choices for the pair of shoes per combination of shirt and skirt:

4 x 2 x 3 = 24 possible outfits

Just before Fred draws the 2nd card, there are 51 cards in the deck, because he already took the 1st card out.

Then once he puts the 2nd card back in, and shuffles, there are still 51 cards in the deck; he did not put the 1st card back in.

So Fred has a 1 in 51 chance of drawing the same card again.

Factorial numbers such as are evaluated as follows;

So to evaluate the expression

(Start)

(Expand the factorials)

(Cancel the common factors)

(Simplify)

This combination problem asks us the number of ways to arrange four letters, including two that are the same. Four different letters can be arranged in  different ways, but since we have two of the same letter, , we have to divide  by . In any combination problem, if we have a total of  letters, then for every  number of the same letters, we have  .

180 is the correct answer. This problem asks for the number of possible permutations of these six letters, which include two letters that are repeated twice,  and . In any combination problem, if we have a total of  letters, then for every  number of the same letters, we have  . So, this problem's situation can be modeled by the expression . We take the factorial of  because we have six letters to be arranged. We divide by  because we have two instances of letters being repeated twice.

Since the four letters are different, they can be arranged in  different ways.