Leah has a list of 8 cities to visit on a road trip with some friends. She needs to finalize the plans for the trip. In how many different orders can she plan to visit the cities?
orders
There are 40,320 different orders Leah can plan to visit the cities.
This can be calculated using the formula for permutations of n objects:
n! = n x (n-1) x (n-2) x ... x 2 x 1
In this case, n = 8 cities.
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320.
So there are 40,320 different orders Leah can plan to visit the cities on her road trip.
Logan is giving an oral presentation on different modes of transportation. He wants to cover buses, horses, and motorcycles. In how many different orders can he include them in his presentation?
orders
There are 6 different orders Logan can include the modes of transportation in his presentation.
This can be calculated using the formula for permutations of n objects:
n! = n x (n-1) x (n-2) x ... x 2 x 1
In this case, n = 3 modes of transportation.
3! = 3 x 2 x 1 = 6.
So there are 6 different orders Logan can include buses, horses, and motorcycles in his presentation.
Some examples of the different orders are:
1. Buses, horses, motorcycles
2. Buses, motorcycles, horses
3. Horses, buses, motorcycles
4. Horses, motorcycles, buses
5. Motorcycles, buses, horses
6. Motorcycles, horses, buses
After moving into a new apartment, Terrell decides to line up mementos on the mantle of the fireplace. If he has 8 mementos to choose from, in how many ways can he pick 2 and line them up?
ways
There are 28 ways Terrell can pick 2 mementos from 8 and line them up.
This can be calculated using the formula for combinations of n objects taken k at a time:
nCk = n! / (k! (n-k)!)
In this case, n = 8 mementos and k = 2 mementos to choose and line up.
8C2 = 8! / (2! (8-2)!)
8C2 = 8! / (2! 6!)
8C2 = (8 x 7 x 6!) / (2 x 1 x 6!)
8C2 = 28
So there are 28 ways Terrell can pick 2 mementos from 8 and line them up on the mantle.
There are 9 employees at a corporate retreat, but a particular team building exercise requires only 2 people, each playing a different role. How many ways are there to select 2 employees and assign them to the roles?
ways
There are 72 ways to select 2 employees and assign them to the roles.
This can be calculated using the formula for combinations of n objects taken k at a time:
nCk = n! / (k! (n-k)!)
In this case, n = 9 employees and k = 2 employees to choose and assign to roles.
9C2 = 9! / (2! (9-2)!)
9C2 = 9! / (2! 7!)
9C2 = (9 x 8 x 7!) / (2 x 1 x 7!)
9C2 = 36
However, we must multiply by the number of ways to assign each selected pair of employees to the two roles. There are 2 options for the first role and 1 option left for the second role. So we multiply 36 by 2 to get the total number of ways to select 2 employees and assign them to the roles:
36 x 2 = 72
Therefore, there are 72 ways to select 2 employees and assign them to the roles for the team building exercise.
A band knows how to play 5 different songs. How many different 2-song set lists could the band perform at their next concert?
set lists
There are 10 different 2-song set lists the band could perform at their next concert.
This can be calculated using the formula for combinations of n objects taken k at a time:
nCk = n! / (k! (n-k)!)
In this case, n = 5 songs and k = 2 songs to choose for the set list.
5C2 = 5! / (2! (5-2)!)
5C2 = 5! / (2! 3!)
5C2 = (5 x 4 x 3!) / (2 x 1 x 3!)
5C2 = 10
So there are 10 different 2-song set lists the band could perform at their next concert.
Another way to think about this is to use the formula for permutations of n objects taken k at a time:
nPk = n! / (n-k)!
In this case, we want to find the number of 2-song set lists, which is the same as arranging 2 songs from 5 in order.
5P2 = 5! / (5-2)!
5P2 = 5! / 3!
5P2 = (5 x 4 x 3!) / 3!
5P2 = 20 / 2
5P2 = 10
So again, there are 10 different 2-song set lists the band could perform at their next concert.
During recess, a child is playing with a bag of 5 marbles, all in different colors. She randomly takes 3 marbles out of the bag, one at a time. How many sequences of 3 marbles are possible?
sequences
There are 60 possible sequences of 3 marbles the child can take out of the bag.
This can be calculated using the formula for permutations of n objects taken k at a time:
nPk = n! / (n-k)!
In this case, n = 5 marbles and k = 3 marbles to take out of the bag in order.
5P3 = 5! / (5-3)!
5P3 = 5! / 2!
5P3 = (5 x 4 x 3 x 2!) / 2!
5P3 = 60
So there are 60 possible sequences of 3 marbles the child can take out of the bag, one at a time.
Cassie owns 2 different toy cribs and she wants to place a single doll in each one. She has 4 dolls to choose from. In how many ways can the cribs be filled?
ways
There are 12 ways that Cassie can put the dolls into the cribs.
Since Cassie can't put two dolls in the same crib, there are 4 options for the first crib and 3 options left for the second crib. So Cassie has 4 x 3 = 12 ways to place the dolls into the cribs.
Therefore, there are 12 ways Cassie can fill the cribs with her dolls.
Clayton owns 3 movies on DVD. He plans to watch a different movie every night for the next 2 nights. In how many ways can he choose which movies to watch and when?
ways
There are 6 ways Clayton can choose which movies to watch and when over the next 2 nights.
There are 3 choices for Clayton to make on the first night: he can choose any one of his 3 movies to watch.
Then there are 2 choices left for Clayton to make on the second night: he can choose from any of the 2 remaining movies.
So, the total number of ways Clayton can choose which movies to watch and when is:
3 choices for the first night x 2 choices for the second night = 6
Therefore, there are 6 ways Clayton can choose which movies to watch and when over the next 2 nights.
There are 4 passengers hoping to get a seat on the next flight to Chicago, but there are only 3 seats left on the plane. In how many ways can the seats be filled?
ways
There are 24 ways that the seats can be filled with the 4 passengers.
This can be calculated using the formula for permutations of n objects taken k at a time:
nPk = n! / (n-k)!
In this case, n = 4 passengers and k = 3 seats to fill.
4P3 = 4! / (4-3)!
4P3 = 4! / 1!
4P3 = 4 x 3 x 2 x 1 / 1
4P3 = 24
So there are 24 ways the seats can be filled with the 4 passengers.
The social action club is choosing members to fill 2 leadership positions. If there are 6 members who are interested in running, in how many different ways can the positions be filled?
ways
There are 15 different ways the leadership positions can be filled by the 6 interested members.
This can be calculated using the formula for combinations of n objects taken k at a time:
nCk = n! / (k! (n-k)!)
In this case, n = 6 interested members and k = 2 positions to fill.
6C2 = 6! / (2! (6-2)!)
6C2 = 6! / (2! 4!)
6C2 = (6 x 5 x 4!) / (2 x 1 x 4!)
6C2 = 15
So there are 15 different ways the 2 leadership positions can be filled by the 6 interested members.
While volunteering on a fundraising campaign, Lee receives a list of 7 people and is asked to call any 2 of them. In how many orders can he select and call 2 people?
orders
There are 21 different ways Lee can select and call 2 people from the list of 7.
This can be calculated using the formula for combinations of n objects taken k at a time:
nCk = n! / (k! (n-k)!)
In this case, n = 7 people and k = 2 people to select and call in any order.
7C2 = 7! / (2! (7-2)!)
7C2 = 7! / (2! 5!)
7C2 = (7 x 6 x 5!) / (2 x 1 x 5!)
7C2 = 21
So there are 21 different orders in which Lee can select and call 2 people from the list of 7.
Sophie has 9 homework assignments due this week, and she needs to finish 2 of them before her friend comes over this afternoon. In how many orders can she pick out and complete 2 of the assignments?
orders
There are 36 different orders in which Sophie can pick out and complete 2 of the assignments.
This can be calculated using the formula for combinations of n objects taken k at a time:
nCk = n! / (k! (n-k)!)
In this case, n = 9 homework assignments and k = 2 assignments to pick and complete in order.
9C2 = 9! / (2! (9-2)!)
9C2 = 9! / (2! 7!)
9C2 = (9 x 8!) / (2 x 1 x 7!)
9C2 = 36
So there are 36 different orders in which Sophie can pick out and complete 2 of the assignments before her friend comes over this afternoon.
A group of 10 students goes to the library to work on homework, but there are only 2 available computers. In how many ways can the students fill the 2 seats at the computers?
ways
There are 45 ways that the students can fill the 2 seats at the computers.
This can be calculated using the formula for combinations of n objects taken k at a time:
nCk = n! / (k! (n-k)!)
In this case, n = 10 students and k = 2 available seats at the computers.
10C2 = 10! / (2! (10-2)!)
10C2 = 10! / (2! 8!)
10C2 = (10 x 9!) / (2 x 1 x 8!)
10C2 = 45
So there are 45 ways the students can fill the 2 seats at the computers.
Isaiah is putting tools into a toolbox, one at a time. He wants to include a crescent wrench, a putty knife, a flathead screwdriver, a pair of pliers, and a hacksaw. In how many different orders could he add these tools?
orders
There are 120 different orders in which Isaiah could add these tools to the toolbox.
This can be calculated using the formula for permutations of n objects:
n! = n x (n-1) x (n-2) x ... x 2 x 1
In this case, n = 5 tools.
5! = 5 x 4 x 3 x 2 x 1 = 120
So there are 120 different orders in which Isaiah could add the crescent wrench, putty knife, flathead screwdriver, pair of pliers, and hacksaw to the toolbox.
The Clark family is taking a photo at their family reunion. There are 2 chairs in the front row and 8 cousins who can sit in them. In how many ways can the cousins be arranged in the seats in the front row?
ways
There are 28 ways that the cousins can be arranged in the 2 seats in the front row.
This can be calculated using the formula for combinations of n objects taken k at a time:
nCk = n! / (k! (n-k)!)
In this case, n = 8 cousins and k = 2 seats in the front row to fill.
8C2 = 8! / (2! (8-2)!)
8C2 = 8! / (2! 6!)
8C2 = (8 x 7) / (2 x 1)
8C2 = 28
So there are 28 ways the cousins can be arranged in the 2 seats in the front row.
Danny owns 5 different blankets. On a particularly cold day, he decides to put 3 blankets on the bed. In how many orders can he pick out the 3 blankets?
orders
There are 60 different orders in which Danny can pick out the 3 blankets.
This can be calculated using the formula for permutations of n objects taken k at a time:
nPk = n! / (n-k)!
In this case, n = 5 blankets and k = 3 blankets to pick out in order.
5P3 = 5! / (5-3)!
5P3 = 5! / 2!
5P3 = (5 x 4 x 3 x 2!) / 2!
5P3 = 60
So there are 60 different orders in which Danny can pick out the 3 blankets to put on the bed.
Marvin is planning a trip to Asia. He has a list of 4 countries he would like to see, but doesn't have time to visit all of them. He asks a travel agent to make an itinerary that includes any 2 of the 4 countries. How many sequences of 2 countries are possible?
sequences
There are 6 different sequences of 2 countries that Marvin can choose from his list of 4 countries.
This can be calculated using the formula for combinations of n objects taken k at a time:
nCk = n! / (k! (n-k)!)
In this case, n = 4 countries to choose from and k = 2 countries to visit.
4C2 = 4! / (2! (4-2)!)
4C2 = 4! / (2! 2!)
4C2 = (4 x 3!) / (2 x 1 x 2!)
4C2 = 6
So there are 6 different sequences of 2 countries that Marvin can choose from his list of 4 countries to include in his itinerary.
Gabrielle is arranging photos for a yearbook page. If she has 5 photos to choose from, in how many ways can she pick 2 of them and line them up on the page?
ways
There are 10 different ways Gabrielle can pick 2 photos and line them up on the page.
This can be calculated using the formula for combinations of n objects taken k at a time:
nCk = n! / (k! (n-k)!)
In this case, n = 5 photos and k = 2 photos to choose and line up.
5C2 = 5! / (2! (5-2)!)
5C2 = 5! / (2! 3!)
5C2 = (5 x 4 x 3!) / (2 x 1 x 3!)
5C2 = 10
So there are 10 different ways Gabrielle can pick 2 photos and line them up on the yearbook page.
After moving into a new house, Cody decides to line up mementos on the mantle of the fireplace. If he has 9 mementos to choose from, in how many ways can he pick 3 and line them up?
ways
There are 84 different ways Cody can pick 3 mementos from 9 and line them up.
This can be calculated using the formula for combinations of n objects taken k at a time:
nCk = n! / (k! (n-k)!)
In this case, n = 9 mementos and k = 3 mementos
Warren has 5 different fish in a small, crowded tank. He decides to randomly scoop up individual fish and transfer them to a new tank until a total of 4 fish have been moved. How many sequences of 4 fish are possible?
sequences
answer?
whats the anser to Warren has 5 different fish in a small, crowded tank. He decides to randomly scoop up individual fish and transfer them to a new tank until a total of 4 fish have been moved. How many sequences of 4 fish are possible?
sequences