The total number of possible combinations of a series of items is the product of the total possibility for each of the items.  Thus, for the letters, there are 26 possibilities for each of the 3 slots, and for the numbers, there are 10 possibilities for each of the 3 slots.  The total number of combinations is then: 26 x 26 x 26 x 10 x 10 x 10 = 17,576,000 ≈ 18 million.

The formula for the number of lines determined by n points, no three of which are “collinear” (on the same line), is n(n-1)/2. To find the number of lines determined by 8 points, we use 8 in the formula to find 8(8-1)/2=8(7)/2=56/2=28. (The formula is derived from two facts: the fact that each point forms a line with each other point, hence n(n-1), and the fact that this relationship is symmetric (i.e. if a forms a line with b, then b forms a line with a), hence dividing by 2.)

The first person holds 7 hands. The second holds six by virtue of already having help the first person’s hand. This continues until through all 8 people. 7+6+5+4+3+2+1=28.

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.