What is the size of the largest set of 10-digit phone numbers such that no two numbers in the set are...
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The problem asks for the largest set of 10-digit numbers which are more different than just by one digit. For example, the set canNOT contain both
$1234567890$
and
$1234527890$
as they have only one different digit.
I've been stuck on this problem for a little while as I can't seem to find any way to approach it.
combinatorics
asked Nov 27 at 5:42
James Radley
1
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The problem asks for the largest set of 10-digit numbers which are more different than just by one digit. For example, the set canNOT contain both
$1234567890$
and
$1234527890$
as they have only one different digit.
I've been stuck on this problem for a little while as I can't seem to find any way to approach it.
combinatorics
asked Nov 27 at 5:42
James Radley
1
Subtract the number of $10$ digit numbers with only one different digit from the total number of $10$ digits possible
– Yadati Kiran
Nov 27 at 6:37
Sure, but how do you count the number of 10 digit numbers with only one different digit from all others?
– James Radley
Nov 28 at 23:23
The first $9$ digits have to be chosen from $10$ digits which is $^{10}C_9$ and the last digit from the chosen $9$ which is $^9C_1$ along with the $10!$ permutations of the $10$ digit number. So you get $10!cdot^{10}C_9cdot^9C_1$.
– Yadati Kiran
Nov 29 at 4:24
This is assuming the phone numbers can start with $0$.
– Yadati Kiran
Nov 29 at 4:34
add a comment |
up vote
0
down vote
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up vote
0
down vote
favorite
The problem asks for the largest set of 10-digit numbers which are more different than just by one digit. For example, the set canNOT contain both
$1234567890$
and
$1234527890$
as they have only one different digit.
I've been stuck on this problem for a little while as I can't seem to find any way to approach it.
combinatorics
asked Nov 27 at 5:42
James Radley
1
The problem asks for the largest set of 10-digit numbers which are more different than just by one digit. For example, the set canNOT contain both
$1234567890$
and
$1234527890$
as they have only one different digit.
I've been stuck on this problem for a little while as I can't seem to find any way to approach it.
combinatorics
combinatorics
asked Nov 27 at 5:42
James Radley
1
asked Nov 27 at 5:42
James Radley
1
asked Nov 27 at 5:42
James Radley
1
asked Nov 27 at 5:42
James Radley
1
asked Nov 27 at 5:42
James Radley
1
1
Subtract the number of $10$ digit numbers with only one different digit from the total number of $10$ digits possible
– Yadati Kiran
Nov 27 at 6:37
Sure, but how do you count the number of 10 digit numbers with only one different digit from all others?
– James Radley
Nov 28 at 23:23
The first $9$ digits have to be chosen from $10$ digits which is $^{10}C_9$ and the last digit from the chosen $9$ which is $^9C_1$ along with the $10!$ permutations of the $10$ digit number. So you get $10!cdot^{10}C_9cdot^9C_1$.
– Yadati Kiran
Nov 29 at 4:24
This is assuming the phone numbers can start with $0$.
– Yadati Kiran
Nov 29 at 4:34
add a comment |
Subtract the number of $10$ digit numbers with only one different digit from the total number of $10$ digits possible
– Yadati Kiran
Nov 27 at 6:37
Sure, but how do you count the number of 10 digit numbers with only one different digit from all others?
– James Radley
Nov 28 at 23:23
The first $9$ digits have to be chosen from $10$ digits which is $^{10}C_9$ and the last digit from the chosen $9$ which is $^9C_1$ along with the $10!$ permutations of the $10$ digit number. So you get $10!cdot^{10}C_9cdot^9C_1$.
– Yadati Kiran
Nov 29 at 4:24
This is assuming the phone numbers can start with $0$.
– Yadati Kiran
Nov 29 at 4:34
Subtract the number of $10$ digit numbers with only one different digit from the total number of $10$ digits possible
– Yadati Kiran
Nov 27 at 6:37
Subtract the number of $10$ digit numbers with only one different digit from the total number of $10$ digits possible
– Yadati Kiran
Nov 27 at 6:37
Sure, but how do you count the number of 10 digit numbers with only one different digit from all others?
– James Radley
Nov 28 at 23:23
Sure, but how do you count the number of 10 digit numbers with only one different digit from all others?
– James Radley
Nov 28 at 23:23
The first $9$ digits have to be chosen from $10$ digits which is $^{10}C_9$ and the last digit from the chosen $9$ which is $^9C_1$ along with the $10!$ permutations of the $10$ digit number. So you get $10!cdot^{10}C_9cdot^9C_1$.
– Yadati Kiran
Nov 29 at 4:24
The first $9$ digits have to be chosen from $10$ digits which is $^{10}C_9$ and the last digit from the chosen $9$ which is $^9C_1$ along with the $10!$ permutations of the $10$ digit number. So you get $10!cdot^{10}C_9cdot^9C_1$.
– Yadati Kiran
Nov 29 at 4:24
This is assuming the phone numbers can start with $0$.
– Yadati Kiran
Nov 29 at 4:34
This is assuming the phone numbers can start with $0$.
– Yadati Kiran
Nov 29 at 4:34
add a comment |
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Subtract the number of $10$ digit numbers with only one different digit from the total number of $10$ digits possible
– Yadati Kiran
Nov 27 at 6:37
Sure, but how do you count the number of 10 digit numbers with only one different digit from all others?
– James Radley
Nov 28 at 23:23
The first $9$ digits have to be chosen from $10$ digits which is $^{10}C_9$ and the last digit from the chosen $9$ which is $^9C_1$ along with the $10!$ permutations of the $10$ digit number. So you get $10!cdot^{10}C_9cdot^9C_1$.
– Yadati Kiran
Nov 29 at 4:24
This is assuming the phone numbers can start with $0$.
– Yadati Kiran
Nov 29 at 4:34