Injective Function Combinations
up vote
1
down vote
favorite
If $|A| = 5$ and $ |B| =25$ , how many functions from $A$ to $B$ are injective?
I'm not quite sure how to tackle this problem as I do not quite understand what $|A| = 5$ and $|B| = 25$ means .
combinatorics discrete-mathematics
edited Nov 24 at 0:21
DeepSea
70.6k54487
asked Nov 24 at 0:07
Charlie
227
add a comment |
up vote
1
down vote
favorite
If $|A| = 5$ and $ |B| =25$ , how many functions from $A$ to $B$ are injective?
I'm not quite sure how to tackle this problem as I do not quite understand what $|A| = 5$ and $|B| = 25$ means .
combinatorics discrete-mathematics
edited Nov 24 at 0:21
DeepSea
70.6k54487
asked Nov 24 at 0:07
Charlie
227
3
$|A|=5$ means $A$ is a set with $5$ elements.
– angryavian
Nov 24 at 0:10
3
$|A|$ is a common notation for the number of elements in the set $|A|$.
– Rob Arthan
Nov 24 at 0:11
Oh ok, thank you!
– Charlie
Nov 24 at 17:24
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $|A| = 5$ and $ |B| =25$ , how many functions from $A$ to $B$ are injective?
I'm not quite sure how to tackle this problem as I do not quite understand what $|A| = 5$ and $|B| = 25$ means .
combinatorics discrete-mathematics
edited Nov 24 at 0:21
DeepSea
70.6k54487
asked Nov 24 at 0:07
Charlie
227
If $|A| = 5$ and $ |B| =25$ , how many functions from $A$ to $B$ are injective?
I'm not quite sure how to tackle this problem as I do not quite understand what $|A| = 5$ and $|B| = 25$ means .
combinatorics discrete-mathematics
combinatorics discrete-mathematics
edited Nov 24 at 0:21
DeepSea
70.6k54487
asked Nov 24 at 0:07
Charlie
227
edited Nov 24 at 0:21
DeepSea
70.6k54487
asked Nov 24 at 0:07
Charlie
227
edited Nov 24 at 0:21
DeepSea
70.6k54487
edited Nov 24 at 0:21
DeepSea
70.6k54487
edited Nov 24 at 0:21
DeepSea
70.6k54487
70.6k54487
asked Nov 24 at 0:07
Charlie
227
asked Nov 24 at 0:07
Charlie
227
asked Nov 24 at 0:07
Charlie
227
227
3
$|A|=5$ means $A$ is a set with $5$ elements.
– angryavian
Nov 24 at 0:10
3
$|A|$ is a common notation for the number of elements in the set $|A|$.
– Rob Arthan
Nov 24 at 0:11
Oh ok, thank you!
– Charlie
Nov 24 at 17:24
add a comment |
3
$|A|=5$ means $A$ is a set with $5$ elements.
– angryavian
Nov 24 at 0:10
3
$|A|$ is a common notation for the number of elements in the set $|A|$.
– Rob Arthan
Nov 24 at 0:11
Oh ok, thank you!
– Charlie
Nov 24 at 17:24
3
3
$|A|=5$ means $A$ is a set with $5$ elements.
– angryavian
Nov 24 at 0:10
$|A|=5$ means $A$ is a set with $5$ elements.
– angryavian
Nov 24 at 0:10
3
3
$|A|$ is a common notation for the number of elements in the set $|A|$.
– Rob Arthan
Nov 24 at 0:11
$|A|$ is a common notation for the number of elements in the set $|A|$.
– Rob Arthan
Nov 24 at 0:11
Oh ok, thank you!
– Charlie
Nov 24 at 17:24
Oh ok, thank you!
– Charlie
Nov 24 at 17:24
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The number of injective maps from $A$ to $B$ is $n!times binom{m}{n}$ with $m = |B|, n = |A|$. In your case, it's $5!times binom{25}{5}$.
answered Nov 24 at 0:18
DeepSea
70.6k54487
Oh ok thanks a lot!
– Charlie
Nov 24 at 17:24
1
@Charlie: Also you can try to find how many surjective maps from $A$ to $B$ with $|A| = 5, |B| = 3$, say.
– DeepSea
Nov 24 at 17:51
add a comment |
up vote
1
down vote
To each elements in A, we associate a unique elements in B, there are $25$ choices for what $1$ gets map to, then $24$ choice for what $2$ gets map to and so on. So $25*24*23*22*21=6 375 600$.
answered Nov 24 at 0:20
mathnoob
1,398116
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The number of injective maps from $A$ to $B$ is $n!times binom{m}{n}$ with $m = |B|, n = |A|$. In your case, it's $5!times binom{25}{5}$.
answered Nov 24 at 0:18
DeepSea
70.6k54487
Oh ok thanks a lot!
– Charlie
Nov 24 at 17:24
1
@Charlie: Also you can try to find how many surjective maps from $A$ to $B$ with $|A| = 5, |B| = 3$, say.
– DeepSea
Nov 24 at 17:51
add a comment |
up vote
1
down vote
accepted
The number of injective maps from $A$ to $B$ is $n!times binom{m}{n}$ with $m = |B|, n = |A|$. In your case, it's $5!times binom{25}{5}$.
answered Nov 24 at 0:18
DeepSea
70.6k54487
Oh ok thanks a lot!
– Charlie
Nov 24 at 17:24
1
@Charlie: Also you can try to find how many surjective maps from $A$ to $B$ with $|A| = 5, |B| = 3$, say.
– DeepSea
Nov 24 at 17:51
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The number of injective maps from $A$ to $B$ is $n!times binom{m}{n}$ with $m = |B|, n = |A|$. In your case, it's $5!times binom{25}{5}$.
answered Nov 24 at 0:18
DeepSea
70.6k54487
The number of injective maps from $A$ to $B$ is $n!times binom{m}{n}$ with $m = |B|, n = |A|$. In your case, it's $5!times binom{25}{5}$.
answered Nov 24 at 0:18
DeepSea
70.6k54487
answered Nov 24 at 0:18
DeepSea
70.6k54487
answered Nov 24 at 0:18
DeepSea
70.6k54487
answered Nov 24 at 0:18
DeepSea
70.6k54487
70.6k54487
Oh ok thanks a lot!
– Charlie
Nov 24 at 17:24
1
@Charlie: Also you can try to find how many surjective maps from $A$ to $B$ with $|A| = 5, |B| = 3$, say.
– DeepSea
Nov 24 at 17:51
add a comment |
Oh ok thanks a lot!
– Charlie
Nov 24 at 17:24
1
@Charlie: Also you can try to find how many surjective maps from $A$ to $B$ with $|A| = 5, |B| = 3$, say.
– DeepSea
Nov 24 at 17:51
Oh ok thanks a lot!
– Charlie
Nov 24 at 17:24
Oh ok thanks a lot!
– Charlie
Nov 24 at 17:24
1
1
@Charlie: Also you can try to find how many surjective maps from $A$ to $B$ with $|A| = 5, |B| = 3$, say.
– DeepSea
Nov 24 at 17:51
@Charlie: Also you can try to find how many surjective maps from $A$ to $B$ with $|A| = 5, |B| = 3$, say.
– DeepSea
Nov 24 at 17:51
add a comment |
up vote
1
down vote
To each elements in A, we associate a unique elements in B, there are $25$ choices for what $1$ gets map to, then $24$ choice for what $2$ gets map to and so on. So $25*24*23*22*21=6 375 600$.
answered Nov 24 at 0:20
mathnoob
1,398116
add a comment |
up vote
1
down vote
To each elements in A, we associate a unique elements in B, there are $25$ choices for what $1$ gets map to, then $24$ choice for what $2$ gets map to and so on. So $25*24*23*22*21=6 375 600$.
answered Nov 24 at 0:20
mathnoob
1,398116
add a comment |
up vote
1
down vote
up vote
1
down vote
To each elements in A, we associate a unique elements in B, there are $25$ choices for what $1$ gets map to, then $24$ choice for what $2$ gets map to and so on. So $25*24*23*22*21=6 375 600$.
answered Nov 24 at 0:20
mathnoob
1,398116
To each elements in A, we associate a unique elements in B, there are $25$ choices for what $1$ gets map to, then $24$ choice for what $2$ gets map to and so on. So $25*24*23*22*21=6 375 600$.
answered Nov 24 at 0:20
mathnoob
1,398116
answered Nov 24 at 0:20
mathnoob
1,398116
answered Nov 24 at 0:20
mathnoob
1,398116
answered Nov 24 at 0:20
mathnoob
1,398116
1,398116
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011001%2finjective-function-combinations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$|A|=5$ means $A$ is a set with $5$ elements.
– angryavian
Nov 24 at 0:10
3
$|A|$ is a common notation for the number of elements in the set $|A|$.
– Rob Arthan
Nov 24 at 0:11
Oh ok, thank you!
– Charlie
Nov 24 at 17:24