He can choose from 10, then 9, then 8 books, but because order does not matter we need to divide by 3 factorial

 (10 * 9 * 8) ÷ (3 * 2 * 1) = 720/6 = 120

Begin by considering the three "hard and fast conditions" - the digits and the one upper-case letter.  For the first number, you will have 10 choices and for the second 9 (since you cannot repeat).  For the captial letter, you have 26 choices.  Thus far, your password has 10 * 9 * 26 possible combinations.

Now, given your remaining options, you have 8 digits, 25 upper-case letters, and 26 lower-case letters (i.e. 59 possible choices).  Since you cannot repeat, you will thus have for your remaining choices 59, 58, and 57 possibilities.

Putting all of this together, you have: 10 * 9 * 26 *59 * 58 * 57 or 456426360 choices.

Using the Fundamental Counting Principle, there would be 8 choices for the first player, 7 choices for the second player, 6 for the third, 5 for the fourth, and so on.  Thus, 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 or 8! = 40, 320.

Since the order of persons shaking hands does not matter, this is a case of computing combinations.  (i.e. It is the same thing for person 1 to shake hands with person 2 as it is for person 2 to shake hands with person 1.)

According to our combinations formula, we have:

300! / ((300-2)! * 2!) = 300! / (298! * 2) = 300 * 299 / 2 = 150 * 299 = 44,850 different handshakes

This is a permutation of 26 letters taken 4 at a time.  To compute this we multiply 26 * 25 * 24 * 23 = 358,800.

The first seat can be filled in 10 ways, the second in 9 ways, the third in 8 ways, and the fourth in 7 ways.  So the number of arrangements = 10 * 9 * 8 * 7 = 5040.

With permutations and combinations, you have to know if the order people are selected matters or not. If not, like in this case, you must take the number of people and positions available:  and divide by number of spots open 

The customer has 3 choices on meat, 6 choices on side, and 5 choices on drink. This gives a total of  choices for the meal.

To solve this problem, you would need to utilize the combination formula, which is .

 is the number of possibilities,  is the number of students, and  is the students attending the meeting. Thus, .

336 would be the result of computing the permutation, which would be incorrect in this case.

This problem involves the permutation of 10 items across 5 slots. The first slot (room) can have 10 different paintings, the second slot can have 9 (one is already in the first room), the third slot can have 8 and so on. The number of possibilities is obtained by multiplying the number of possible options in each of the 5 slots together, which here is .