Permutation / Combination - SAT Math
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If there are 8 points in a plane, and no 3 of the points lie along the same line, how many unique lines can be drawn between pairs of these 8 points?
If there are 8 points in a plane, and no 3 of the points lie along the same line, how many unique lines can be drawn between pairs of these 8 points?
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 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.)
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8 people locked in a room take turns holding hands with each person only once. How many hand holdings take place?
8 people locked in a room take turns holding hands with each person only once. How many hand holdings take place?
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.
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.
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If a team of 7 basketball players win a game and each high-five each other once after a game, how many high-fives take place?
If a team of 7 basketball players win a game and each high-five each other once after a game, how many high-fives take place?
To start, the first player makes 6 high-fives. The second player needs to make only 5 new high-fives because he doesn't need to re high-five the first player. The third player needs to only make 4 high-fives and so on.
This adds up to 6 + 5 + 4 + 3 + 2 + 1 = 21
To start, the first player makes 6 high-fives. The second player needs to make only 5 new high-fives because he doesn't need to re high-five the first player. The third player needs to only make 4 high-fives and so on.
This adds up to 6 + 5 + 4 + 3 + 2 + 1 = 21
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Mr. and Mrs. Williams have invited three other married couples to a dinner party; the eight are to be seated at the table shown above. Since Mr. and Mrs. Williams need to be close to the kitchen so that they can serve the meal, it is desired that they sit at Seats 1 and 8, although it does not matter which one sits in which seat. There are no other restrictions.
How many ways can the eight persons be seated to fit this specification?
Mr. and Mrs. Williams have invited three other married couples to a dinner party; the eight are to be seated at the table shown above. Since Mr. and Mrs. Williams need to be close to the kitchen so that they can serve the meal, it is desired that they sit at Seats 1 and 8, although it does not matter which one sits in which seat. There are no other restrictions.
How many ways can the eight persons be seated to fit this specification?
Since Seats 1 and 8 will be occupied by the Williamses, there are two ways to fill these seats - Mr. Williams in Seat 1 and Mrs. Williams in Seat 8, and the opposite case.
The remaining six persons will be seated in the other six seats; since order is material here, this is a number of permutations. The number of permutations of six elements from a set of six is
,
which is equal to
.
By the multiplication principle, the number of seating arrangements is
.
Since Seats 1 and 8 will be occupied by the Williamses, there are two ways to fill these seats - Mr. Williams in Seat 1 and Mrs. Williams in Seat 8, and the opposite case.
The remaining six persons will be seated in the other six seats; since order is material here, this is a number of permutations. The number of permutations of six elements from a set of six is
which is equal to
By the multiplication principle, the number of seating arrangements is
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Mr. and Mrs. Jones have invited three other married couples to a dinner party; the eight are to be seated at the table shown above. It is desired that each husband sit directly across from his wife -for example, a husband at Seat 1, his wife at Seat 6.
How many ways can the eight persons be seated to fit this specification?
Mr. and Mrs. Jones have invited three other married couples to a dinner party; the eight are to be seated at the table shown above. It is desired that each husband sit directly across from his wife -for example, a husband at Seat 1, his wife at Seat 6.
How many ways can the eight persons be seated to fit this specification?
There are four pairs of seats directly across from each other: 1-6, 2-5, 3-8, 4-7.
There are four married couples to be placed in these four pairs of seats; since order is relevant, these are permutations. The number of ways to seat these couples is equal to the number of orderings, or permutations, of four elements from a set of four:
,
which is equal to
Within each pairing, there are two ways for the husband to select one seat and the wife to select the other, and there are four such pairings, so, applying the multiplication principle, the number of arrangements is
There are four pairs of seats directly across from each other: 1-6, 2-5, 3-8, 4-7.
There are four married couples to be placed in these four pairs of seats; since order is relevant, these are permutations. The number of ways to seat these couples is equal to the number of orderings, or permutations, of four elements from a set of four:
which is equal to
Within each pairing, there are two ways for the husband to select one seat and the wife to select the other, and there are four such pairings, so, applying the multiplication principle, the number of arrangements is
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Yvonne wants to invite seven of her twenty classmates to her birthday party. She and her classmates will all be seated at the above table. Which of the following expressions gives the number of ways she can select seven classmates and seat them, and herself, at the table?
Yvonne wants to invite seven of her twenty classmates to her birthday party. She and her classmates will all be seated at the above table. Which of the following expressions gives the number of ways she can select seven classmates and seat them, and herself, at the table?
One way to look at this is as follows:
Yvonne can sit at any one of the eight seats—eight ways to make this decision.
Yvonne can fill the remaining seven seats with seven out of twenty classmates. Since order is important, this is a permutation of seven elements chosen from a set of twenty; there are 
ways to make this decision.
By the multiplication principle, there are 
ways Yvonne can invite seven classmates and seat them and herself at the table.
One way to look at this is as follows:
Yvonne can sit at any one of the eight seats—eight ways to make this decision.
Yvonne can fill the remaining seven seats with seven out of twenty classmates. Since order is important, this is a permutation of seven elements chosen from a set of twenty; there are
By the multiplication principle, there are
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There are seven unique placemats around a circular table. How many different orders of placemats are possible?
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.
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.
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Mark has 5 pants and 7 shirts in his closet. He wants to wear a different pant/shirt combination each day without buying new clothes for as long as he can. How many weeks can he do this for?
Mark has 5 pants and 7 shirts in his closet. He wants to wear a different pant/shirt combination each day without buying new clothes for as long as he can. How many weeks can he do this for?
The fundamental counting principle says that if you want to determine the number of ways that two independent events can happen, multiply the number of ways each event can happen together. In this case, there are 5 * 7, or 35 unique combinations of pants & shirts Mark can wear. If he wears one combination each day, he can last 35 days, or 5 weeks, without buying new clothes.
The fundamental counting principle says that if you want to determine the number of ways that two independent events can happen, multiply the number of ways each event can happen together. In this case, there are 5 * 7, or 35 unique combinations of pants & shirts Mark can wear. If he wears one combination each day, he can last 35 days, or 5 weeks, without buying new clothes.
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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?
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?
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.
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.
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Twenty students enter a contest at school. The contest offers a first, second, and third prize. How many different combinations of 1st, 2nd, and 3rd place winners can there be?
Twenty students enter a contest at school. The contest offers a first, second, and third prize. How many different combinations of 1st, 2nd, and 3rd place winners can there be?
This is a permutation problem, because we are looking for the number of groups of winners. Consider the three positions, and how many choices there are for each position: There are 20 choices for 1st place, 19 for 2nd place, and 18 for 3rd place.
20, 19, 18
Multiply to get 6840.
This is a permutation problem, because we are looking for the number of groups of winners. Consider the three positions, and how many choices there are for each position: There are 20 choices for 1st place, 19 for 2nd place, and 18 for 3rd place.
20, 19, 18
Multiply to get 6840.
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A baker has four different types of frosting, three different kinds of sprinkles, and 8 different cookie cutters. How many different cookie combinations can the baker create if each cookie has one type of frosting and one type of sprinkle?
A baker has four different types of frosting, three different kinds of sprinkles, and 8 different cookie cutters. How many different cookie combinations can the baker create if each cookie has one type of frosting and one type of sprinkle?
Since this a combination problem and we want to know how many different ways the cookies can be created we can solve this using the Fundamental counting principle. 4 x 3 x 8 = 96
Multiplying each of the possible choices together.
Since this a combination problem and we want to know how many different ways the cookies can be created we can solve this using the Fundamental counting principle. 4 x 3 x 8 = 96
Multiplying each of the possible choices together.
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If a series of license plates is to be produced that all have the same pattern of three letters followed by three numbers, roughly how many alphanumeric combinations are possible?
If a series of license plates is to be produced that all have the same pattern of three letters followed by three numbers, roughly how many alphanumeric combinations are possible?
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 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.
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At an ice cream store, there are 5 flavors of ice cream: strawberry, vanilla, chocolate, mint, and banana. How many different 3-flavor ice cream cones can be made?
At an ice cream store, there are 5 flavors of ice cream: strawberry, vanilla, chocolate, mint, and banana. How many different 3-flavor ice cream cones can be made?
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: nCr = 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.)
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: nCr = 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.)
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At a deli you can choose from either Italian bread, whole wheat bread, or sourdough bread. You can choose turkey or roast beef as your meat and provolone or mozzarella as your cheese. If you have to choose a bread, a meat, and a cheese, how many possible sandwich combinations can you have?
At a deli you can choose from either Italian bread, whole wheat bread, or sourdough bread. You can choose turkey or roast beef as your meat and provolone or mozzarella as your cheese. If you have to choose a bread, a meat, and a cheese, how many possible sandwich combinations can you have?
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.
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.
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Shannon decided to go to nearby café for lunch. She can have a sandwich made on either wheat or white bread. The café offers cheddar, Swiss, and American for cheese choices. For meat, Shannon can choose ham, turkey, bologna, roast beef, or salami. How many cheese and meat sandwich options does Shannon have to choose from?
Shannon decided to go to nearby café for lunch. She can have a sandwich made on either wheat or white bread. The café offers cheddar, Swiss, and American for cheese choices. For meat, Shannon can choose ham, turkey, bologna, roast beef, or salami. How many cheese and meat sandwich options does Shannon have to choose from?
2 bread choices * 3 cheese choices * 5 meat choices = 30 sandwich choices
2 bread choices * 3 cheese choices * 5 meat choices = 30 sandwich choices
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An ice cream parlor serves 36 ice cream flavors. You can order any flavor in a small, medium or large and can choose between a waffle cone and a cup. How many possible combinations could you possibly order?
An ice cream parlor serves 36 ice cream flavors. You can order any flavor in a small, medium or large and can choose between a waffle cone and a cup. How many possible combinations could you possibly order?
36 possible flavors * 3 possible sizes * 2 possible cones = 216 possible combinations.
36 possible flavors * 3 possible sizes * 2 possible cones = 216 possible combinations.
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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?
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?
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 _S_1, _S_2, and _S_3. We will call the four entrees _E_1, _E_2, _E_3, and _E_4, and we will call the desserts _D_1 and _D_2.
Here are the possible lunch special combinations:
_S_1, _E_1, _D_1
_S_1, _E_1, _D_2
_S_1, _E_2, _D_1
_S_1, _E_2, _D_2
_S_1, _E_3, _D_1
_S_1, _E_3, _D_2
_S_1, _E_4, _D_1
_S_1, _E_4, _D_2
_S_2, _E_1, _D_1
_S_2, _E_1, _D_2
_S_2, _E_2, _D_1
_S_2, _E_2, _D_2
_S_2, _E_3, _D_1
_S_2, _E_3, _D_2
_S_2, _E_4, _D_1
_S_2, _E_4, _D_2
_S_3, _E_1, _D_1
_S_3, _E_1, _D_2
_S_3, _E_2, _D_1
_S_3, _E_2, _D_2
_S_3, _E_3, _D_1
_S_3, _E_3, _D_2
_S_3, _E_4, _D_1
_S_3, _E_4, _D_2
The answer is 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 _S_1, _S_2, and _S_3. We will call the four entrees _E_1, _E_2, _E_3, and _E_4, and we will call the desserts _D_1 and _D_2.
Here are the possible lunch special combinations:
_S_1, _E_1, _D_1
_S_1, _E_1, _D_2
_S_1, _E_2, _D_1
_S_1, _E_2, _D_2
_S_1, _E_3, _D_1
_S_1, _E_3, _D_2
_S_1, _E_4, _D_1
_S_1, _E_4, _D_2
_S_2, _E_1, _D_1
_S_2, _E_1, _D_2
_S_2, _E_2, _D_1
_S_2, _E_2, _D_2
_S_2, _E_3, _D_1
_S_2, _E_3, _D_2
_S_2, _E_4, _D_1
_S_2, _E_4, _D_2
_S_3, _E_1, _D_1
_S_3, _E_1, _D_2
_S_3, _E_2, _D_1
_S_3, _E_2, _D_2
_S_3, _E_3, _D_1
_S_3, _E_3, _D_2
_S_3, _E_4, _D_1
_S_3, _E_4, _D_2
The answer is 24.
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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?
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.
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
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
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Ernie has 4 shirts, 7 ties, and 3 pants. How many different possible outfits can Ernie make?
Ernie has 4 shirts, 7 ties, and 3 pants. How many different possible outfits can Ernie make?
The purpose of this question is to understand how to calculate a number of combinations. All different combinations of clothing articles are equally possible. Since all combinations are possible, the numbers of articles of clothing are all multiplied together, 
, yielding 84 different possible outfits.
The purpose of this question is to understand how to calculate a number of combinations. All different combinations of clothing articles are equally possible. Since all combinations are possible, the numbers of articles of clothing are all multiplied together,
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