Combinatorics - How many numbers with all different digits are between $1000$ and $7856$?
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To answer this question I was trying to solve the problem by parts.
So, first I counted from 1000 to 7000: $6*9*8*7$ that means there are $6$ possibilities on position one, $9$ to position 2, $8$ to position 3 and $7$ for the last one.
Then, I counted from 7000 to 7800 and I got this: $7*8*7$.
In tens, I counted from 7800 to 7850 and I think this part is wrong but I got this:
$4*7$, when there are 4 possibilities on tens and 7 possibilities on units because I need to put away 3 digits above.
Finally, on units I only got 5, so the all sum is: $6*9*8*7+7*8*7+4*7+5=3449$.
I want to know if I solve this problem well. I'm learning English, so I'm sorry if you can't understand me.
combinatorics combinations
asked Aug 13 at 3:52
Johnny Villegas
111
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up vote
2
down vote
favorite
To answer this question I was trying to solve the problem by parts.
So, first I counted from 1000 to 7000: $6*9*8*7$ that means there are $6$ possibilities on position one, $9$ to position 2, $8$ to position 3 and $7$ for the last one.
Then, I counted from 7000 to 7800 and I got this: $7*8*7$.
In tens, I counted from 7800 to 7850 and I think this part is wrong but I got this:
$4*7$, when there are 4 possibilities on tens and 7 possibilities on units because I need to put away 3 digits above.
Finally, on units I only got 5, so the all sum is: $6*9*8*7+7*8*7+4*7+5=3449$.
I want to know if I solve this problem well. I'm learning English, so I'm sorry if you can't understand me.
combinatorics combinations
asked Aug 13 at 3:52
Johnny Villegas
111
3
You have the right idea, but you should be careful about how you phrase things. "I counted from 7000 to 7800" sounds like you were counting $7000leq x leq 7800$ when in reality you should have split it up as $7000leq x < 7800$. Although not relevant for that particular case, it becomes relevant for the next two cases. That being said, in counting the number of distinct digit numbers $x$ where $7800leq x < 7850$, you in fact have five options for the tens digit, not four, giving $5cdot 7$ numbers in this case.
– JMoravitz
Aug 13 at 3:59
1
As for the final case, that depends on whether "between 1000 and 7856" is meant to mean $1000leq x leq 7856$ (between inclusively) or if it is meant to mean $1000< x < 7856$ (between exclusively). If inclusive, then there are in fact six possibilities, not five.
– JMoravitz
Aug 13 at 4:00
Hi and thanks for your answers! I figured out. First of all, I'm working with the inclusively interval, I mean $1000≤x≤7856$ but when I taken all the splitted intervals I work without including the last part of the interval like that $7000≤x<7800$. Thank you guys again and have a nice day!
– Johnny Villegas
Aug 14 at 12:49
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
To answer this question I was trying to solve the problem by parts.
So, first I counted from 1000 to 7000: $6*9*8*7$ that means there are $6$ possibilities on position one, $9$ to position 2, $8$ to position 3 and $7$ for the last one.
Then, I counted from 7000 to 7800 and I got this: $7*8*7$.
In tens, I counted from 7800 to 7850 and I think this part is wrong but I got this:
$4*7$, when there are 4 possibilities on tens and 7 possibilities on units because I need to put away 3 digits above.
Finally, on units I only got 5, so the all sum is: $6*9*8*7+7*8*7+4*7+5=3449$.
I want to know if I solve this problem well. I'm learning English, so I'm sorry if you can't understand me.
combinatorics combinations
asked Aug 13 at 3:52
Johnny Villegas
111
To answer this question I was trying to solve the problem by parts.
So, first I counted from 1000 to 7000: $6*9*8*7$ that means there are $6$ possibilities on position one, $9$ to position 2, $8$ to position 3 and $7$ for the last one.
Then, I counted from 7000 to 7800 and I got this: $7*8*7$.
In tens, I counted from 7800 to 7850 and I think this part is wrong but I got this:
$4*7$, when there are 4 possibilities on tens and 7 possibilities on units because I need to put away 3 digits above.
Finally, on units I only got 5, so the all sum is: $6*9*8*7+7*8*7+4*7+5=3449$.
I want to know if I solve this problem well. I'm learning English, so I'm sorry if you can't understand me.
combinatorics combinations
asked Aug 13 at 3:52
Johnny Villegas
111
asked Aug 13 at 3:52
Johnny Villegas
111
asked Aug 13 at 3:52
Johnny Villegas
111
asked Aug 13 at 3:52
Johnny Villegas
111
111
3
You have the right idea, but you should be careful about how you phrase things. "I counted from 7000 to 7800" sounds like you were counting $7000leq x leq 7800$ when in reality you should have split it up as $7000leq x < 7800$. Although not relevant for that particular case, it becomes relevant for the next two cases. That being said, in counting the number of distinct digit numbers $x$ where $7800leq x < 7850$, you in fact have five options for the tens digit, not four, giving $5cdot 7$ numbers in this case.
– JMoravitz
Aug 13 at 3:59
1
As for the final case, that depends on whether "between 1000 and 7856" is meant to mean $1000leq x leq 7856$ (between inclusively) or if it is meant to mean $1000< x < 7856$ (between exclusively). If inclusive, then there are in fact six possibilities, not five.
– JMoravitz
Aug 13 at 4:00
Hi and thanks for your answers! I figured out. First of all, I'm working with the inclusively interval, I mean $1000≤x≤7856$ but when I taken all the splitted intervals I work without including the last part of the interval like that $7000≤x<7800$. Thank you guys again and have a nice day!
– Johnny Villegas
Aug 14 at 12:49
add a comment |Â
3
You have the right idea, but you should be careful about how you phrase things. "I counted from 7000 to 7800" sounds like you were counting $7000leq x leq 7800$ when in reality you should have split it up as $7000leq x < 7800$. Although not relevant for that particular case, it becomes relevant for the next two cases. That being said, in counting the number of distinct digit numbers $x$ where $7800leq x < 7850$, you in fact have five options for the tens digit, not four, giving $5cdot 7$ numbers in this case.
– JMoravitz
Aug 13 at 3:59
1
As for the final case, that depends on whether "between 1000 and 7856" is meant to mean $1000leq x leq 7856$ (between inclusively) or if it is meant to mean $1000< x < 7856$ (between exclusively). If inclusive, then there are in fact six possibilities, not five.
– JMoravitz
Aug 13 at 4:00
Hi and thanks for your answers! I figured out. First of all, I'm working with the inclusively interval, I mean $1000≤x≤7856$ but when I taken all the splitted intervals I work without including the last part of the interval like that $7000≤x<7800$. Thank you guys again and have a nice day!
– Johnny Villegas
Aug 14 at 12:49
3
3
You have the right idea, but you should be careful about how you phrase things. "I counted from 7000 to 7800" sounds like you were counting $7000leq x leq 7800$ when in reality you should have split it up as $7000leq x < 7800$. Although not relevant for that particular case, it becomes relevant for the next two cases. That being said, in counting the number of distinct digit numbers $x$ where $7800leq x < 7850$, you in fact have five options for the tens digit, not four, giving $5cdot 7$ numbers in this case.
– JMoravitz
Aug 13 at 3:59
You have the right idea, but you should be careful about how you phrase things. "I counted from 7000 to 7800" sounds like you were counting $7000leq x leq 7800$ when in reality you should have split it up as $7000leq x < 7800$. Although not relevant for that particular case, it becomes relevant for the next two cases. That being said, in counting the number of distinct digit numbers $x$ where $7800leq x < 7850$, you in fact have five options for the tens digit, not four, giving $5cdot 7$ numbers in this case.
– JMoravitz
Aug 13 at 3:59
1
1
As for the final case, that depends on whether "between 1000 and 7856" is meant to mean $1000leq x leq 7856$ (between inclusively) or if it is meant to mean $1000< x < 7856$ (between exclusively). If inclusive, then there are in fact six possibilities, not five.
– JMoravitz
Aug 13 at 4:00
As for the final case, that depends on whether "between 1000 and 7856" is meant to mean $1000leq x leq 7856$ (between inclusively) or if it is meant to mean $1000< x < 7856$ (between exclusively). If inclusive, then there are in fact six possibilities, not five.
– JMoravitz
Aug 13 at 4:00
Hi and thanks for your answers! I figured out. First of all, I'm working with the inclusively interval, I mean $1000≤x≤7856$ but when I taken all the splitted intervals I work without including the last part of the interval like that $7000≤x<7800$. Thank you guys again and have a nice day!
– Johnny Villegas
Aug 14 at 12:49
Hi and thanks for your answers! I figured out. First of all, I'm working with the inclusively interval, I mean $1000≤x≤7856$ but when I taken all the splitted intervals I work without including the last part of the interval like that $7000≤x<7800$. Thank you guys again and have a nice day!
– Johnny Villegas
Aug 14 at 12:49
add a comment |Â
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3
You have the right idea, but you should be careful about how you phrase things. "I counted from 7000 to 7800" sounds like you were counting $7000leq x leq 7800$ when in reality you should have split it up as $7000leq x < 7800$. Although not relevant for that particular case, it becomes relevant for the next two cases. That being said, in counting the number of distinct digit numbers $x$ where $7800leq x < 7850$, you in fact have five options for the tens digit, not four, giving $5cdot 7$ numbers in this case.
– JMoravitz
Aug 13 at 3:59
1
As for the final case, that depends on whether "between 1000 and 7856" is meant to mean $1000leq x leq 7856$ (between inclusively) or if it is meant to mean $1000< x < 7856$ (between exclusively). If inclusive, then there are in fact six possibilities, not five.
– JMoravitz
Aug 13 at 4:00
Hi and thanks for your answers! I figured out. First of all, I'm working with the inclusively interval, I mean $1000≤x≤7856$ but when I taken all the splitted intervals I work without including the last part of the interval like that $7000≤x<7800$. Thank you guys again and have a nice day!
– Johnny Villegas
Aug 14 at 12:49