Roger is spending the afternoon house hunting. There are 4 properties on his list, and he has time to see 2 of them. How many sequences of 2 houses are possible?
sequences
There are 6 possible sequences of 2 houses for Roger to see.
To see why, we can use the formula for combinations:
nCr = n! / r!(n-r)!
In this case, n = 4 (the number of properties on his list) and r = 2 (the number of houses he will see).
4C2 = 4! / 2!(4-2)! = 6
So there are 6 possible sequences of 2 houses for Roger to see. They are:
- 1st and 2nd houses
- 1st and 3rd houses
- 1st and 4th houses
- 2nd and 3rd houses
- 2nd and 4th houses
- 3rd and 4th houses
At a dog show, a Pekingese, a Chihuahua, and a Great Dane won first place in their divisions. In how many different orders could these winners be announced?
orders
There are 6 different orders in which these winners could be announced.
To see why, we can list out all the possible orders:
1. Pekingese, Chihuahua, Great Dane
2. Pekingese, Great Dane, Chihuahua
3. Chihuahua, Pekingese, Great Dane
4. Chihuahua, Great Dane, Pekingese
5. Great Dane, Pekingese, Chihuahua
6. Great Dane, Chihuahua, Pekingese
So there are 6 different orders in which the winners could be announced.
There are 8 projects that made it to the final round of the science competition. In how many different orders can the judges rank these projects?
orders
There are 40,320 different orders in which the judges can rank these projects.
To see why, we can use the formula for permutations:
n! = n(n-1)(n-2)...(3)(2)(1)
In this case, n = 8 (the number of projects in the final round).
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 in which the judges can rank these projects.
Rose has 10 books on her shelf, and she wants to read any 2 of them on vacation. How many sequences of 2 books are possible?
sequences
There are 45 sequences of 2 books that Rose can choose from her shelf.
To see why, we can use the formula for combinations:
nCr = n! / r!(n-r)!
In this case, n = 10 (the number of books on her shelf) and r = 2 (the number of books she will read on vacation).
10C2 = 10! / 2!(10-2)! = 45
So there are 45 sequences of 2 books that Rose can choose from her shelf.
Kevin is hanging framed photos above his fireplace. If he has 3 photos to choose from and 2 hooks on which to hang them, how many arrangements are possible?
ways
There are 6 different arrangements that are possible for Kevin to hang his 3 photos on the 2 hooks.
To see why, we can list out all the possible arrangements:
1. Photo 1 on hook 1, Photo 2 on hook 2, Photo 3 not hung
2. Photo 1 on hook 1, Photo 3 on hook 2, Photo 2 not hung
3. Photo 2 on hook 1, Photo 1 on hook 2, Photo 3 not hung
4. Photo 2 on hook 1, Photo 3 on hook 2, Photo 1 not hung
5. Photo 3 on hook 1, Photo 1 on hook 2, Photo 2 not hung
6. Photo 3 on hook 1, Photo 2 on hook 2, Photo 1 not hung
So there are 6 different arrangements that are possible for Kevin to hang his 3 photos on the 2 hooks.