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Example Questions
Example Question #841 : Arithmetic
Example Question #331 : Arithmetic
A group of friends decide to go out to the movies. Fred and Tom are bringing dates, while their 2 friends are going alone. When the friends arrive at the movie theater, they find a row of six seats so they can all sit together.
If Fred and Tom must sit next to their dates, how many different ways can the group sit down?
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.
Example Question #11 : How To Find The Greatest Or Least Number Of Combinations
There are seven unique placemats around a circular table. How many different orders of placemats are possible?
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.
Example Question #12 : How To Find The Greatest Or Least Number Of Combinations
For a certain lunch special, customers must order a salad, an entree, and a dessert. If there are three different salads, four different entrees, and two different desserts available, then how many different lunch specials are possible?
18
9
12
24
72
24
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.
Example Question #13 : How To Find The Greatest Or Least Number Of Combinations
There are five pictures but only four display cases. The display cases are unique. How many different arrangements of pictures in display cases can be created?
5
240
24
50
120
120
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.
Example Question #1892 : Psat Mathematics
The art club must choose a leadership committee of 3 students. If any
member can be on the committee, how many different combnations of 3
students can be selected from the 15 members of the art club?
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