Coloring Red and Blue Problem












-1














We color the integers from 1 to 999 with red and blue, so that each integer is assigned one of the two colors. How many different colorings can we construct with the property that there are more red integers within the numbers {1,...,500} than within the numbers {501,...999}






combinatorics coloring







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  • 2




    Please edit your question to show what you have attempted. For instance, have you solved the problem of determining how many ways we can color the integers from $1$ to $9$ in such a way that there are more red integers within the set ${1, 2, 3, 4, 5}$ than there are in the set ${6, 7, 8, 9}$?
    – N. F. Taussig
    Dec 2 at 18:06










  • Hint: Can you see that the number of colorings with more reds among ${1,dots,500}$ than reds among ${501,dots,599}$ is the same as the number of colorings with more reds among ${1,dots,500}$ than blues among ${501,dots,599}$?
    – bof
    Dec 3 at 11:16










  • Hint: Can you see that the condition "more reds among ${1,dots,500}$ than blues among ${501,dots,599}$" is equivalent to "more reds than blues overall"?
    – bof
    Dec 3 at 11:18












  • How many ways can you color the integers from $1$ to $999$ with red and blue, so that there are more reds than blues?
    – bof
    Dec 3 at 11:19
















-1














We color the integers from 1 to 999 with red and blue, so that each integer is assigned one of the two colors. How many different colorings can we construct with the property that there are more red integers within the numbers {1,...,500} than within the numbers {501,...999}






combinatorics coloring







share|cite|improve this question


















  • 2




    Please edit your question to show what you have attempted. For instance, have you solved the problem of determining how many ways we can color the integers from $1$ to $9$ in such a way that there are more red integers within the set ${1, 2, 3, 4, 5}$ than there are in the set ${6, 7, 8, 9}$?
    – N. F. Taussig
    Dec 2 at 18:06










  • Hint: Can you see that the number of colorings with more reds among ${1,dots,500}$ than reds among ${501,dots,599}$ is the same as the number of colorings with more reds among ${1,dots,500}$ than blues among ${501,dots,599}$?
    – bof
    Dec 3 at 11:16










  • Hint: Can you see that the condition "more reds among ${1,dots,500}$ than blues among ${501,dots,599}$" is equivalent to "more reds than blues overall"?
    – bof
    Dec 3 at 11:18












  • How many ways can you color the integers from $1$ to $999$ with red and blue, so that there are more reds than blues?
    – bof
    Dec 3 at 11:19














-1












-1








-1







We color the integers from 1 to 999 with red and blue, so that each integer is assigned one of the two colors. How many different colorings can we construct with the property that there are more red integers within the numbers {1,...,500} than within the numbers {501,...999}






combinatorics coloring







share|cite|improve this question













We color the integers from 1 to 999 with red and blue, so that each integer is assigned one of the two colors. How many different colorings can we construct with the property that there are more red integers within the numbers {1,...,500} than within the numbers {501,...999}






combinatorics coloring




combinatorics coloring






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 at 17:18









Bob Johnson

1




1








  • 2




    Please edit your question to show what you have attempted. For instance, have you solved the problem of determining how many ways we can color the integers from $1$ to $9$ in such a way that there are more red integers within the set ${1, 2, 3, 4, 5}$ than there are in the set ${6, 7, 8, 9}$?
    – N. F. Taussig
    Dec 2 at 18:06










  • Hint: Can you see that the number of colorings with more reds among ${1,dots,500}$ than reds among ${501,dots,599}$ is the same as the number of colorings with more reds among ${1,dots,500}$ than blues among ${501,dots,599}$?
    – bof
    Dec 3 at 11:16










  • Hint: Can you see that the condition "more reds among ${1,dots,500}$ than blues among ${501,dots,599}$" is equivalent to "more reds than blues overall"?
    – bof
    Dec 3 at 11:18












  • How many ways can you color the integers from $1$ to $999$ with red and blue, so that there are more reds than blues?
    – bof
    Dec 3 at 11:19














  • 2




    Please edit your question to show what you have attempted. For instance, have you solved the problem of determining how many ways we can color the integers from $1$ to $9$ in such a way that there are more red integers within the set ${1, 2, 3, 4, 5}$ than there are in the set ${6, 7, 8, 9}$?
    – N. F. Taussig
    Dec 2 at 18:06










  • Hint: Can you see that the number of colorings with more reds among ${1,dots,500}$ than reds among ${501,dots,599}$ is the same as the number of colorings with more reds among ${1,dots,500}$ than blues among ${501,dots,599}$?
    – bof
    Dec 3 at 11:16










  • Hint: Can you see that the condition "more reds among ${1,dots,500}$ than blues among ${501,dots,599}$" is equivalent to "more reds than blues overall"?
    – bof
    Dec 3 at 11:18












  • How many ways can you color the integers from $1$ to $999$ with red and blue, so that there are more reds than blues?
    – bof
    Dec 3 at 11:19








2




2




Please edit your question to show what you have attempted. For instance, have you solved the problem of determining how many ways we can color the integers from $1$ to $9$ in such a way that there are more red integers within the set ${1, 2, 3, 4, 5}$ than there are in the set ${6, 7, 8, 9}$?
– N. F. Taussig
Dec 2 at 18:06




Please edit your question to show what you have attempted. For instance, have you solved the problem of determining how many ways we can color the integers from $1$ to $9$ in such a way that there are more red integers within the set ${1, 2, 3, 4, 5}$ than there are in the set ${6, 7, 8, 9}$?
– N. F. Taussig
Dec 2 at 18:06












Hint: Can you see that the number of colorings with more reds among ${1,dots,500}$ than reds among ${501,dots,599}$ is the same as the number of colorings with more reds among ${1,dots,500}$ than blues among ${501,dots,599}$?
– bof
Dec 3 at 11:16




Hint: Can you see that the number of colorings with more reds among ${1,dots,500}$ than reds among ${501,dots,599}$ is the same as the number of colorings with more reds among ${1,dots,500}$ than blues among ${501,dots,599}$?
– bof
Dec 3 at 11:16












Hint: Can you see that the condition "more reds among ${1,dots,500}$ than blues among ${501,dots,599}$" is equivalent to "more reds than blues overall"?
– bof
Dec 3 at 11:18






Hint: Can you see that the condition "more reds among ${1,dots,500}$ than blues among ${501,dots,599}$" is equivalent to "more reds than blues overall"?
– bof
Dec 3 at 11:18














How many ways can you color the integers from $1$ to $999$ with red and blue, so that there are more reds than blues?
– bof
Dec 3 at 11:19




How many ways can you color the integers from $1$ to $999$ with red and blue, so that there are more reds than blues?
– bof
Dec 3 at 11:19










1 Answer
1






active

oldest

votes


















-1














First you can chose how many red numbers there will be in ${1,..,500}$

Suppose you chose to color $k$ red numbers in that set. ($k$ can take any value from 0 to 500)

In the ${501,..,999}$ set, you can color from $0$ to $k-1$ numbers in red (k possibilities)



Thus if we iterate over every possible value for $k$ you have $sum_{k=0}^{500}k^2$ distinct possiblilites

We can generalize for a set of length $n$: {$1,..,n$} and {$n+1,.., 2n-1$}

There are a total of $sum_{k=0}^nk^2=frac{n(n+1)(2n+1)}{6}$ possiblilites






share|cite|improve this answer





















  • This answer doesn't take into account the process of actually choosing $k$ numbers from 1 to 500, and similar for the other set.
    – platty
    Dec 3 at 10:42











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1














First you can chose how many red numbers there will be in ${1,..,500}$

Suppose you chose to color $k$ red numbers in that set. ($k$ can take any value from 0 to 500)

In the ${501,..,999}$ set, you can color from $0$ to $k-1$ numbers in red (k possibilities)



Thus if we iterate over every possible value for $k$ you have $sum_{k=0}^{500}k^2$ distinct possiblilites

We can generalize for a set of length $n$: {$1,..,n$} and {$n+1,.., 2n-1$}

There are a total of $sum_{k=0}^nk^2=frac{n(n+1)(2n+1)}{6}$ possiblilites






share|cite|improve this answer





















  • This answer doesn't take into account the process of actually choosing $k$ numbers from 1 to 500, and similar for the other set.
    – platty
    Dec 3 at 10:42
















-1














First you can chose how many red numbers there will be in ${1,..,500}$

Suppose you chose to color $k$ red numbers in that set. ($k$ can take any value from 0 to 500)

In the ${501,..,999}$ set, you can color from $0$ to $k-1$ numbers in red (k possibilities)



Thus if we iterate over every possible value for $k$ you have $sum_{k=0}^{500}k^2$ distinct possiblilites

We can generalize for a set of length $n$: {$1,..,n$} and {$n+1,.., 2n-1$}

There are a total of $sum_{k=0}^nk^2=frac{n(n+1)(2n+1)}{6}$ possiblilites






share|cite|improve this answer





















  • This answer doesn't take into account the process of actually choosing $k$ numbers from 1 to 500, and similar for the other set.
    – platty
    Dec 3 at 10:42














-1












-1








-1






First you can chose how many red numbers there will be in ${1,..,500}$

Suppose you chose to color $k$ red numbers in that set. ($k$ can take any value from 0 to 500)

In the ${501,..,999}$ set, you can color from $0$ to $k-1$ numbers in red (k possibilities)



Thus if we iterate over every possible value for $k$ you have $sum_{k=0}^{500}k^2$ distinct possiblilites

We can generalize for a set of length $n$: {$1,..,n$} and {$n+1,.., 2n-1$}

There are a total of $sum_{k=0}^nk^2=frac{n(n+1)(2n+1)}{6}$ possiblilites






share|cite|improve this answer












First you can chose how many red numbers there will be in ${1,..,500}$

Suppose you chose to color $k$ red numbers in that set. ($k$ can take any value from 0 to 500)

In the ${501,..,999}$ set, you can color from $0$ to $k-1$ numbers in red (k possibilities)



Thus if we iterate over every possible value for $k$ you have $sum_{k=0}^{500}k^2$ distinct possiblilites

We can generalize for a set of length $n$: {$1,..,n$} and {$n+1,.., 2n-1$}

There are a total of $sum_{k=0}^nk^2=frac{n(n+1)(2n+1)}{6}$ possiblilites







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 3 at 10:27









TheD0ubleT

39218




39218












  • This answer doesn't take into account the process of actually choosing $k$ numbers from 1 to 500, and similar for the other set.
    – platty
    Dec 3 at 10:42


















  • This answer doesn't take into account the process of actually choosing $k$ numbers from 1 to 500, and similar for the other set.
    – platty
    Dec 3 at 10:42
















This answer doesn't take into account the process of actually choosing $k$ numbers from 1 to 500, and similar for the other set.
– platty
Dec 3 at 10:42




This answer doesn't take into account the process of actually choosing $k$ numbers from 1 to 500, and similar for the other set.
– platty
Dec 3 at 10:42


















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