How many integers between $1000$ and $9999$ inclusive have distinct digits and are odd numbers?
$begingroup$
I know this question was already answered (How many integers between 1000 and 9999 inclusive consist of), and I understand the solution. However, I don't understand what's wrong with my approach. I did it from right to left. My reasoning:
Last digit has to be odd, so we have $5$ possibilities. In the tenths place, we can choose whatever we want except for the one already chosen, so we have $9$ possibilities. Similarly, in the hundredths place we have $8$ choices. In the thousands place, we can choose $10 - 3 - 1$ (not including $0$) $= 6$ possibilities, so in total we have $5 * 9 * 8 * 6 = 2160$. Why am I undercounting here?
combinatorics
asked Dec 2 '18 at 18:37
Dumb DumbDumb Dumb
363
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add a comment |
$begingroup$
I know this question was already answered (How many integers between 1000 and 9999 inclusive consist of), and I understand the solution. However, I don't understand what's wrong with my approach. I did it from right to left. My reasoning:
Last digit has to be odd, so we have $5$ possibilities. In the tenths place, we can choose whatever we want except for the one already chosen, so we have $9$ possibilities. Similarly, in the hundredths place we have $8$ choices. In the thousands place, we can choose $10 - 3 - 1$ (not including $0$) $= 6$ possibilities, so in total we have $5 * 9 * 8 * 6 = 2160$. Why am I undercounting here?
combinatorics
asked Dec 2 '18 at 18:37
Dumb DumbDumb Dumb
363
$endgroup$
add a comment |
$begingroup$
I know this question was already answered (How many integers between 1000 and 9999 inclusive consist of), and I understand the solution. However, I don't understand what's wrong with my approach. I did it from right to left. My reasoning:
Last digit has to be odd, so we have $5$ possibilities. In the tenths place, we can choose whatever we want except for the one already chosen, so we have $9$ possibilities. Similarly, in the hundredths place we have $8$ choices. In the thousands place, we can choose $10 - 3 - 1$ (not including $0$) $= 6$ possibilities, so in total we have $5 * 9 * 8 * 6 = 2160$. Why am I undercounting here?
combinatorics
asked Dec 2 '18 at 18:37
Dumb DumbDumb Dumb
363
$endgroup$
I know this question was already answered (How many integers between 1000 and 9999 inclusive consist of), and I understand the solution. However, I don't understand what's wrong with my approach. I did it from right to left. My reasoning:
Last digit has to be odd, so we have $5$ possibilities. In the tenths place, we can choose whatever we want except for the one already chosen, so we have $9$ possibilities. Similarly, in the hundredths place we have $8$ choices. In the thousands place, we can choose $10 - 3 - 1$ (not including $0$) $= 6$ possibilities, so in total we have $5 * 9 * 8 * 6 = 2160$. Why am I undercounting here?
combinatorics
combinatorics
asked Dec 2 '18 at 18:37
Dumb DumbDumb Dumb
363
asked Dec 2 '18 at 18:37
Dumb DumbDumb Dumb
363
asked Dec 2 '18 at 18:37
Dumb DumbDumb Dumb
363
asked Dec 2 '18 at 18:37
Dumb DumbDumb Dumb
363
asked Dec 2 '18 at 18:37
Dumb DumbDumb Dumb
363
363
add a comment |
add a comment |
1 Answer
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$begingroup$
Because if you've already chosen $0$ before (in the tens or hundreds place), then you're excluding $0$ twice in allowing only $10-3-1=6$ options for the thousands place.
To compensate this, you could count the number of cases where you have a $0$ in the tens or hundreds place. But as in the linked answer, it is easier to just start with the thousands place.
answered Dec 2 '18 at 18:41
ServaesServaes
22.5k33793
$endgroup$
$begingroup$
I see. Thank you!
$endgroup$
– Dumb Dumb
Dec 2 '18 at 19:51
add a comment |
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1 Answer
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oldest
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$begingroup$
Because if you've already chosen $0$ before (in the tens or hundreds place), then you're excluding $0$ twice in allowing only $10-3-1=6$ options for the thousands place.
To compensate this, you could count the number of cases where you have a $0$ in the tens or hundreds place. But as in the linked answer, it is easier to just start with the thousands place.
answered Dec 2 '18 at 18:41
ServaesServaes
22.5k33793
$endgroup$
$begingroup$
I see. Thank you!
$endgroup$
– Dumb Dumb
Dec 2 '18 at 19:51
add a comment |
$begingroup$
Because if you've already chosen $0$ before (in the tens or hundreds place), then you're excluding $0$ twice in allowing only $10-3-1=6$ options for the thousands place.
To compensate this, you could count the number of cases where you have a $0$ in the tens or hundreds place. But as in the linked answer, it is easier to just start with the thousands place.
answered Dec 2 '18 at 18:41
ServaesServaes
22.5k33793
$endgroup$
$begingroup$
I see. Thank you!
$endgroup$
– Dumb Dumb
Dec 2 '18 at 19:51
add a comment |
$begingroup$
Because if you've already chosen $0$ before (in the tens or hundreds place), then you're excluding $0$ twice in allowing only $10-3-1=6$ options for the thousands place.
To compensate this, you could count the number of cases where you have a $0$ in the tens or hundreds place. But as in the linked answer, it is easier to just start with the thousands place.
answered Dec 2 '18 at 18:41
ServaesServaes
22.5k33793
$endgroup$
Because if you've already chosen $0$ before (in the tens or hundreds place), then you're excluding $0$ twice in allowing only $10-3-1=6$ options for the thousands place.
To compensate this, you could count the number of cases where you have a $0$ in the tens or hundreds place. But as in the linked answer, it is easier to just start with the thousands place.
answered Dec 2 '18 at 18:41
ServaesServaes
22.5k33793
answered Dec 2 '18 at 18:41
ServaesServaes
22.5k33793
answered Dec 2 '18 at 18:41
ServaesServaes
22.5k33793
answered Dec 2 '18 at 18:41
ServaesServaes
22.5k33793
22.5k33793
$begingroup$
I see. Thank you!
$endgroup$
– Dumb Dumb
Dec 2 '18 at 19:51
add a comment |
$begingroup$
I see. Thank you!
$endgroup$
– Dumb Dumb
Dec 2 '18 at 19:51
$begingroup$
I see. Thank you!
$endgroup$
– Dumb Dumb
Dec 2 '18 at 19:51
$begingroup$
I see. Thank you!
$endgroup$
– Dumb Dumb
Dec 2 '18 at 19:51
add a comment |
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