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

This is a combinations problem which means order does not matter. For his first choice the president can choose from 6, the second 5, and the third 4 so you may think the answer is 6 * 5 * 4, or 120; however this would be the answer if he were choosing an ordered set like vice president, secretary of state, and chief of staff. In this case order does not matter, so you must divide the 6 * 5 * 4 by 3 * 2 * 1, for the three seats he’s choosing. The answer is 120/6, or 20.

Because order is not important in this problem (i.e. chocolate chip, pecan, butterscotch is no different than pecan, butterscotch, chocolate chip), it is dealing with combinations rather than permutations.

The formula for a combination is given as:

where  is the number of options and  is the size of the combination.

For the ice cream choices, there are thirty-one options to build a three-scoop sundae. So, the number of ice cream combinations is given as:

Now, for the topping combinations, we are told there are ten options and that Mohammed is allowed to pick two items; however, we are also told that Mohammed has already chosen one, so this leaves nine options with one item being selected:

So there are 9 "combinations" (using the word a bit loosely) available for the toppings. This is perhaps intuitive, but it's worth doing the math.

To solve this problem, we must understand the concept of combination/permutations. A combination is used when the order doesn't matter while a permutation is used when order matters. In this problem, the two class representatives are randomly chosen, therefore it doesn't matter what order the representative is chosen in, the end result is the same.  The general formula for combinations is , where  is the number of things you have and  is the things you want to combine.

Plugging in choosing 2 people from a group of 20, we find

.  Therefore there are a  different ways to choose the  class representatives.

The number of potential combinations for  selections made from  possible options is

Quantity A:

Quantity B:

Quantity A is greater.

For  choices made from  possible options, the number of potential combinations (order does not matter) is

And the number of potential permutations (order matters) is

Quantity A:

Quantity B:

Quantity A is greater.

When dealing with combinations, the number of possible combinations when selecting  choices out of  options is:

For Quantity A, the number of combinations is:

For Quantity B, the number of combinations is:

Quantity B is greater.