There are 5x4x3 ways to arrange 5 flavors in 3 ways.  However, in this case, the order of the flavors does not matter (e.g., a cone with strawberry, mint, and banana is the same as a cone with mint, banana, and strawberry).  So we have to divide 5x4x3 by the number of ways we can arrange 3 different things which is 3x2x1.  So (5x4x3)/(3x2x1) is 10.  

One can also use the combination formula for this problem: _nC_r = n! / (n-r)! r!

Therefore: ₅C₃ = 5! / 3! 2!

= 10

(Note: an example of a counting problem in which order would matter is a lock or passcode situation.  The permutation 3-5-7 for a three number lock or passcode is a distinct outcome from 5-7-3, and thus both must be counted.)

You have 3 possible types of bread, 2 possible types of meat, and 2 possible types of cheese. Multiplying them out you get 3*2*2, giving you 12 possible combinations.

2 bread choices * 3 cheese choices * 5 meat choices = 30 sandwich choices

36 possible flavors * 3 possible sizes * 2 possible cones = 216 possible combinations.

Think of the seats as an arrangement of people in a line.  Fred and Tom must sit next to their dates, so you can treat the pair as a single object.  The only difference is that we must then multiply by 2, since we can switch the order in which they sit down at will (either Fred or his date can sit on the left).

So instead of dealing with 6 objects, we now simply work with 4.  An arrangement of 4 objects, can be made in different ways.  You can choose any of 4 objects to be in the first spot. Once that spot is taken, you move onto the next of four spots.  You place any of the remaining three there, giving you 3 more choices (or multiplying by 3).  You do the same thing 2 more times to end up with 24 possibilities.

Finally, you have to take into account switching the positions of Fred/Tom and their respective dates.  Since there are two pairs, you multiply by 2 twice.  This gives you 

different arrangements.

Since the table is circular, you need to find the total number of orders and divide this number by 7.

The total number of different orders that the placemats could be set in is 7! (7 factorial).

7!/7 = 6! = 720

Note that had this been a linear, and not circular, arrangement there would be no need to divide by 7. But in a circular arrangement there are no "ends" so you must divide by N! by N to account for the circular arrangement.

Customers must choose a salad, an entree, and a dessert. There are three different salads, four entrees, and two desserts.

The simplest way of determining the number of combinations is by multiplying the number of options for each part of the meal. In other words, we can find the product of 3, 4, and 2, which would give us 24.

Sometimes, if you can't think of a way to mathetimatically determine all of the different combinations of something, it helps to write out as many as you can. Let's write out all of the possible cominbations just to verify that there are 24. Let's call the different salads S1, S2, and S3. We will call the four entrees E1, E2, E3, and E4, and we will call the desserts D1 and D2.

Here are the possible lunch special combinations:

S1, E1, D1

S1, E1, D2

S1, E2, D1

S1, E2, D2

S1, E3, D1

S1, E3, D2

S1, E4, D1

S1, E4, D2

S2, E1, D1

S2, E1, D2

S2, E2, D1

S2, E2, D2

S2, E3, D1

S2, E3, D2

S2, E4, D1

S2, E4, D2

S3, E1, D1

S3, E1, D2

S3, E2, D1

S3, E2, D2

S3, E3, D1

S3, E3, D2

S3, E4, D1

S3, E4, D2

The answer is 24.

There are five possible choices for the first space. For the second there are four possible, three for the third, and two for the fourth. 5 * 4 * 3 * 2 = 120 possible arrangements.