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 .

There are  different permutations of 3 people from a group of seven (when order matters). There are  possible ways to arrange 3 people. Thus when order doesn't matter, there can be  different committees formed. 

Although this is a permutation style problem, we have to be careful regarding the last portion (i.e. the side dishes).  We know that our meal will have:

(3 possible salads) * (5 possible sandwiches) * (x possible combinations of side dishes).

We must ascertain how many side dishes can be selected.  Now, it does not matter what order we put together the side dishes, so we have to use the combinations formula:

c(n,r) = (n!) / ((n-r)! * (r!))

Plugging in, we get: c(15,2) = 15! / (13! * 2!) = 15 * 14 / 2 = 15 * 7 = 105

Using this in the equation above, we get: 3 * 5 * 105 = 1575