Combinatorics problem Algebra
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Four friends have four different balls and they decide to swap them so that everyone has a different ball. How many possible combinations are there?
I know they are $9$: $2, 1, 4, 3 - 2, 3, 4, 1 - 2, 4, 1, 3 - 3, 1, 4, 2 -
3, 4, 1, 2 - 3, 4, 2, 1 - 4, 1, 2, 3 - 4, 3, 1, 2 - 4, 3, 2, 1$ but how can i calculate them without writing them down?
combinatorics
asked Nov 18 at 10:08
Sdwae
52
add a comment |
up vote
0
down vote
favorite
Four friends have four different balls and they decide to swap them so that everyone has a different ball. How many possible combinations are there?
I know they are $9$: $2, 1, 4, 3 - 2, 3, 4, 1 - 2, 4, 1, 3 - 3, 1, 4, 2 -
3, 4, 1, 2 - 3, 4, 2, 1 - 4, 1, 2, 3 - 4, 3, 1, 2 - 4, 3, 2, 1$ but how can i calculate them without writing them down?
combinatorics
asked Nov 18 at 10:08
Sdwae
52
You mean noone has his own ball?
– greedoid
Nov 18 at 10:11
Yes, not the one that he had at the start
– Sdwae
Nov 18 at 10:13
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Four friends have four different balls and they decide to swap them so that everyone has a different ball. How many possible combinations are there?
I know they are $9$: $2, 1, 4, 3 - 2, 3, 4, 1 - 2, 4, 1, 3 - 3, 1, 4, 2 -
3, 4, 1, 2 - 3, 4, 2, 1 - 4, 1, 2, 3 - 4, 3, 1, 2 - 4, 3, 2, 1$ but how can i calculate them without writing them down?
combinatorics
asked Nov 18 at 10:08
Sdwae
52
Four friends have four different balls and they decide to swap them so that everyone has a different ball. How many possible combinations are there?
I know they are $9$: $2, 1, 4, 3 - 2, 3, 4, 1 - 2, 4, 1, 3 - 3, 1, 4, 2 -
3, 4, 1, 2 - 3, 4, 2, 1 - 4, 1, 2, 3 - 4, 3, 1, 2 - 4, 3, 2, 1$ but how can i calculate them without writing them down?
combinatorics
combinatorics
asked Nov 18 at 10:08
Sdwae
52
asked Nov 18 at 10:08
Sdwae
52
asked Nov 18 at 10:08
Sdwae
52
asked Nov 18 at 10:08
Sdwae
52
asked Nov 18 at 10:08
Sdwae
52
52
You mean noone has his own ball?
– greedoid
Nov 18 at 10:11
Yes, not the one that he had at the start
– Sdwae
Nov 18 at 10:13
add a comment |
You mean noone has his own ball?
– greedoid
Nov 18 at 10:11
Yes, not the one that he had at the start
– Sdwae
Nov 18 at 10:13
You mean noone has his own ball?
– greedoid
Nov 18 at 10:11
You mean noone has his own ball?
– greedoid
Nov 18 at 10:11
Yes, not the one that he had at the start
– Sdwae
Nov 18 at 10:13
Yes, not the one that he had at the start
– Sdwae
Nov 18 at 10:13
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
These are called derangements and have been well studied. For any $n$ there are $$
n!Sigma_{i=0}^nfrac{(-1)^i}{i!}
$$
answered Nov 18 at 10:11
Joey Kilpatrick
1,163321
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
These are called derangements and have been well studied. For any $n$ there are $$
n!Sigma_{i=0}^nfrac{(-1)^i}{i!}
$$
answered Nov 18 at 10:11
Joey Kilpatrick
1,163321
add a comment |
up vote
1
down vote
accepted
These are called derangements and have been well studied. For any $n$ there are $$
n!Sigma_{i=0}^nfrac{(-1)^i}{i!}
$$
answered Nov 18 at 10:11
Joey Kilpatrick
1,163321
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
These are called derangements and have been well studied. For any $n$ there are $$
n!Sigma_{i=0}^nfrac{(-1)^i}{i!}
$$
answered Nov 18 at 10:11
Joey Kilpatrick
1,163321
These are called derangements and have been well studied. For any $n$ there are $$
n!Sigma_{i=0}^nfrac{(-1)^i}{i!}
$$
answered Nov 18 at 10:11
Joey Kilpatrick
1,163321
answered Nov 18 at 10:11
Joey Kilpatrick
1,163321
answered Nov 18 at 10:11
Joey Kilpatrick
1,163321
answered Nov 18 at 10:11
Joey Kilpatrick
1,163321
1,163321
add a comment |
add a comment |
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You mean noone has his own ball?
– greedoid
Nov 18 at 10:11
Yes, not the one that he had at the start
– Sdwae
Nov 18 at 10:13