Mixing
Count number of functions
Clash Royale CLAN TAG#URR8PPP
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Given $A=biglfmid f:1,2,3 rightarrow 1,2,3,4bigr,; f$ is a function.
How many elements of $A$ satisfy $f(3)>f(2)$?
I know that the total amount of functions is $4^3$, I am not sure how to proceed with the sets needed. Can you help me define the sets I need to work with?
combinatorics
edited Jul 31 at 16:30
N. F. Taussig
38k93053
asked Jul 31 at 14:17
Jorge Chang
254
add a comment |Â
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Given $A=biglfmid f:1,2,3 rightarrow 1,2,3,4bigr,; f$ is a function.
How many elements of $A$ satisfy $f(3)>f(2)$?
I know that the total amount of functions is $4^3$, I am not sure how to proceed with the sets needed. Can you help me define the sets I need to work with?
combinatorics
edited Jul 31 at 16:30
N. F. Taussig
38k93053
asked Jul 31 at 14:17
Jorge Chang
254
Note that there's a preview under the textbox where you enter the question, and an edit button under the question once you've submitted it, so you can still fix the syntax error that's messing up the first line. (Note also that MathJax uses braces for grouping, so even when you've balanced them correctly, you'll need to escape them with backslashes (...) to make them visible.)
– joriki
Jul 31 at 14:23
Thank you and sorry.
– Jorge Chang
Jul 31 at 14:28
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given $A=biglfmid f:1,2,3 rightarrow 1,2,3,4bigr,; f$ is a function.
How many elements of $A$ satisfy $f(3)>f(2)$?
I know that the total amount of functions is $4^3$, I am not sure how to proceed with the sets needed. Can you help me define the sets I need to work with?
combinatorics
edited Jul 31 at 16:30
N. F. Taussig
38k93053
asked Jul 31 at 14:17
Jorge Chang
254
Given $A=biglfmid f:1,2,3 rightarrow 1,2,3,4bigr,; f$ is a function.
How many elements of $A$ satisfy $f(3)>f(2)$?
I know that the total amount of functions is $4^3$, I am not sure how to proceed with the sets needed. Can you help me define the sets I need to work with?
combinatorics
edited Jul 31 at 16:30
N. F. Taussig
38k93053
asked Jul 31 at 14:17
Jorge Chang
254
edited Jul 31 at 16:30
N. F. Taussig
38k93053
edited Jul 31 at 16:30
N. F. Taussig
38k93053
edited Jul 31 at 16:30
N. F. Taussig
38k93053
38k93053
asked Jul 31 at 14:17
Jorge Chang
254
asked Jul 31 at 14:17
Jorge Chang
254
asked Jul 31 at 14:17
Jorge Chang
254
254
Note that there's a preview under the textbox where you enter the question, and an edit button under the question once you've submitted it, so you can still fix the syntax error that's messing up the first line. (Note also that MathJax uses braces for grouping, so even when you've balanced them correctly, you'll need to escape them with backslashes (...) to make them visible.)
– joriki
Jul 31 at 14:23
Thank you and sorry.
– Jorge Chang
Jul 31 at 14:28
add a comment |Â
Note that there's a preview under the textbox where you enter the question, and an edit button under the question once you've submitted it, so you can still fix the syntax error that's messing up the first line. (Note also that MathJax uses braces for grouping, so even when you've balanced them correctly, you'll need to escape them with backslashes (...) to make them visible.)
– joriki
Jul 31 at 14:23
Thank you and sorry.
– Jorge Chang
Jul 31 at 14:28
Note that there's a preview under the textbox where you enter the question, and an edit button under the question once you've submitted it, so you can still fix the syntax error that's messing up the first line. (Note also that MathJax uses braces for grouping, so even when you've balanced them correctly, you'll need to escape them with backslashes (
...) to make them visible.)– joriki
Jul 31 at 14:23
Note that there's a preview under the textbox where you enter the question, and an edit button under the question once you've submitted it, so you can still fix the syntax error that's messing up the first line. (Note also that MathJax uses braces for grouping, so even when you've balanced them correctly, you'll need to escape them with backslashes (
...) to make them visible.)– joriki
Jul 31 at 14:23
Thank you and sorry.
– Jorge Chang
Jul 31 at 14:28
Thank you and sorry.
– Jorge Chang
Jul 31 at 14:28
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Hint: Each two-element subset of $1,2,3,4$ can be uniquely assigned to the values of $f(2),f(3)$ such that $f(2)<f(3)$ and vice versa. For example $2,4$ would correspond with $f(2)=2$ and $f(3)=4$, while $4,1$ would correspond with $f(2)=1$ and $f(3)=4$. (Remember $4,1=1,4$, that order within a /set/ is irrelevant)
How many ways are there to choose a two-element subset of a four-element set? This is then the number of ways of selecting the values for $f(2)$ and $f(3)$. Couple this as well with the number of ways of selecting the value of $f(1)$ and apply the rule of product to arrive at a final answer.
Your answer in the (now deleted) comments of the other answer: "I think I got it. Is it possible to add 4(3)+4(2)+4(1)? First choosing the element for f(2) and counting the numbers that make f(3)>f(2)?" This is correct and $4(3)+4(2)+4(1)$ is indeed the correct answer... However... it feels a bit too much like brute force and does not as easily generalize as it otherwise could. What if we were asking for $f(3)<f(4)<f(5)<dots<f(100)$ and our codomain were $1,2,3,4,dots,1000$. Your method would not have been feasible, but the method I am trying to lead you to above is still straightforward and easy to calculate.
answered Jul 31 at 14:48
JMoravitz
44k33481
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: Each two-element subset of $1,2,3,4$ can be uniquely assigned to the values of $f(2),f(3)$ such that $f(2)<f(3)$ and vice versa. For example $2,4$ would correspond with $f(2)=2$ and $f(3)=4$, while $4,1$ would correspond with $f(2)=1$ and $f(3)=4$. (Remember $4,1=1,4$, that order within a /set/ is irrelevant)
How many ways are there to choose a two-element subset of a four-element set? This is then the number of ways of selecting the values for $f(2)$ and $f(3)$. Couple this as well with the number of ways of selecting the value of $f(1)$ and apply the rule of product to arrive at a final answer.
Your answer in the (now deleted) comments of the other answer: "I think I got it. Is it possible to add 4(3)+4(2)+4(1)? First choosing the element for f(2) and counting the numbers that make f(3)>f(2)?" This is correct and $4(3)+4(2)+4(1)$ is indeed the correct answer... However... it feels a bit too much like brute force and does not as easily generalize as it otherwise could. What if we were asking for $f(3)<f(4)<f(5)<dots<f(100)$ and our codomain were $1,2,3,4,dots,1000$. Your method would not have been feasible, but the method I am trying to lead you to above is still straightforward and easy to calculate.
answered Jul 31 at 14:48
JMoravitz
44k33481
add a comment |Â
up vote
2
down vote
accepted
Hint: Each two-element subset of $1,2,3,4$ can be uniquely assigned to the values of $f(2),f(3)$ such that $f(2)<f(3)$ and vice versa. For example $2,4$ would correspond with $f(2)=2$ and $f(3)=4$, while $4,1$ would correspond with $f(2)=1$ and $f(3)=4$. (Remember $4,1=1,4$, that order within a /set/ is irrelevant)
How many ways are there to choose a two-element subset of a four-element set? This is then the number of ways of selecting the values for $f(2)$ and $f(3)$. Couple this as well with the number of ways of selecting the value of $f(1)$ and apply the rule of product to arrive at a final answer.
Your answer in the (now deleted) comments of the other answer: "I think I got it. Is it possible to add 4(3)+4(2)+4(1)? First choosing the element for f(2) and counting the numbers that make f(3)>f(2)?" This is correct and $4(3)+4(2)+4(1)$ is indeed the correct answer... However... it feels a bit too much like brute force and does not as easily generalize as it otherwise could. What if we were asking for $f(3)<f(4)<f(5)<dots<f(100)$ and our codomain were $1,2,3,4,dots,1000$. Your method would not have been feasible, but the method I am trying to lead you to above is still straightforward and easy to calculate.
answered Jul 31 at 14:48
JMoravitz
44k33481
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: Each two-element subset of $1,2,3,4$ can be uniquely assigned to the values of $f(2),f(3)$ such that $f(2)<f(3)$ and vice versa. For example $2,4$ would correspond with $f(2)=2$ and $f(3)=4$, while $4,1$ would correspond with $f(2)=1$ and $f(3)=4$. (Remember $4,1=1,4$, that order within a /set/ is irrelevant)
How many ways are there to choose a two-element subset of a four-element set? This is then the number of ways of selecting the values for $f(2)$ and $f(3)$. Couple this as well with the number of ways of selecting the value of $f(1)$ and apply the rule of product to arrive at a final answer.
Your answer in the (now deleted) comments of the other answer: "I think I got it. Is it possible to add 4(3)+4(2)+4(1)? First choosing the element for f(2) and counting the numbers that make f(3)>f(2)?" This is correct and $4(3)+4(2)+4(1)$ is indeed the correct answer... However... it feels a bit too much like brute force and does not as easily generalize as it otherwise could. What if we were asking for $f(3)<f(4)<f(5)<dots<f(100)$ and our codomain were $1,2,3,4,dots,1000$. Your method would not have been feasible, but the method I am trying to lead you to above is still straightforward and easy to calculate.
answered Jul 31 at 14:48
JMoravitz
44k33481
Hint: Each two-element subset of $1,2,3,4$ can be uniquely assigned to the values of $f(2),f(3)$ such that $f(2)<f(3)$ and vice versa. For example $2,4$ would correspond with $f(2)=2$ and $f(3)=4$, while $4,1$ would correspond with $f(2)=1$ and $f(3)=4$. (Remember $4,1=1,4$, that order within a /set/ is irrelevant)
How many ways are there to choose a two-element subset of a four-element set? This is then the number of ways of selecting the values for $f(2)$ and $f(3)$. Couple this as well with the number of ways of selecting the value of $f(1)$ and apply the rule of product to arrive at a final answer.
Your answer in the (now deleted) comments of the other answer: "I think I got it. Is it possible to add 4(3)+4(2)+4(1)? First choosing the element for f(2) and counting the numbers that make f(3)>f(2)?" This is correct and $4(3)+4(2)+4(1)$ is indeed the correct answer... However... it feels a bit too much like brute force and does not as easily generalize as it otherwise could. What if we were asking for $f(3)<f(4)<f(5)<dots<f(100)$ and our codomain were $1,2,3,4,dots,1000$. Your method would not have been feasible, but the method I am trying to lead you to above is still straightforward and easy to calculate.
answered Jul 31 at 14:48
JMoravitz
44k33481
answered Jul 31 at 14:48
JMoravitz
44k33481
answered Jul 31 at 14:48
JMoravitz
44k33481
answered Jul 31 at 14:48
JMoravitz
44k33481
44k33481
add a comment |Â
add a comment |Â
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Note that there's a preview under the textbox where you enter the question, and an edit button under the question once you've submitted it, so you can still fix the syntax error that's messing up the first line. (Note also that MathJax uses braces for grouping, so even when you've balanced them correctly, you'll need to escape them with backslashes (
...) to make them visible.)– joriki
Jul 31 at 14:23
Thank you and sorry.
– Jorge Chang
Jul 31 at 14:28