All 6-digit numbers consist of [6] which must contain 1,2,3
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I'm required to find all possible numbers which consist of {1,2,3,4,5,6}, and include each of {1,2,3} at least once. For example: 113124 is good, but not 464612 because it's missing 3.
My approach was to first calculate all the sequences which contain 1,2,3 once. Which would be: $6times5times4times3^3$. The reasoning is that you have 6 different places to place 3, then 5 for 2, and 4 for 1. Because I'm looking now for sequences which contain each digit only once I'd fill the 3 rest spots with {4,5,6} meaning $3^3$. Now for the case where you have a digit twice it would be: $3times{6choose2}times4times3times3^2$. Using the same reasoning but this time you have $6choose2$ ways to choose 2 positions for a number, then you are left with 4 places for the other number and so on. And you multiply by 3 for each number. So you carry on up to 4 times the same number, because if you would continue for 5 you don't have enough spots to fill in the other 2 numbers you must have, so the formula is:
$$ 6*5*4*3^3 + 3*{6choose2}*4*3*3^2+3*{6choose3}*3*2*3+3*{6choose4}*2$$
Yet the answer is incorret, it's a question a friend asked me to do so I'm not aware of the exact solution, from what I recall he said the answer is roughly 11,000.
combinatorics discrete-mathematics
asked Jan 14 at 10:40
MosheMoshe
396
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add a comment |
$begingroup$
I'm required to find all possible numbers which consist of {1,2,3,4,5,6}, and include each of {1,2,3} at least once. For example: 113124 is good, but not 464612 because it's missing 3.
My approach was to first calculate all the sequences which contain 1,2,3 once. Which would be: $6times5times4times3^3$. The reasoning is that you have 6 different places to place 3, then 5 for 2, and 4 for 1. Because I'm looking now for sequences which contain each digit only once I'd fill the 3 rest spots with {4,5,6} meaning $3^3$. Now for the case where you have a digit twice it would be: $3times{6choose2}times4times3times3^2$. Using the same reasoning but this time you have $6choose2$ ways to choose 2 positions for a number, then you are left with 4 places for the other number and so on. And you multiply by 3 for each number. So you carry on up to 4 times the same number, because if you would continue for 5 you don't have enough spots to fill in the other 2 numbers you must have, so the formula is:
$$ 6*5*4*3^3 + 3*{6choose2}*4*3*3^2+3*{6choose3}*3*2*3+3*{6choose4}*2$$
Yet the answer is incorret, it's a question a friend asked me to do so I'm not aware of the exact solution, from what I recall he said the answer is roughly 11,000.
combinatorics discrete-mathematics
asked Jan 14 at 10:40
MosheMoshe
396
$endgroup$
1
$begingroup$
Inclusion-Exclusion
$endgroup$
– Haran
Jan 14 at 10:48
add a comment |
$begingroup$
I'm required to find all possible numbers which consist of {1,2,3,4,5,6}, and include each of {1,2,3} at least once. For example: 113124 is good, but not 464612 because it's missing 3.
My approach was to first calculate all the sequences which contain 1,2,3 once. Which would be: $6times5times4times3^3$. The reasoning is that you have 6 different places to place 3, then 5 for 2, and 4 for 1. Because I'm looking now for sequences which contain each digit only once I'd fill the 3 rest spots with {4,5,6} meaning $3^3$. Now for the case where you have a digit twice it would be: $3times{6choose2}times4times3times3^2$. Using the same reasoning but this time you have $6choose2$ ways to choose 2 positions for a number, then you are left with 4 places for the other number and so on. And you multiply by 3 for each number. So you carry on up to 4 times the same number, because if you would continue for 5 you don't have enough spots to fill in the other 2 numbers you must have, so the formula is:
$$ 6*5*4*3^3 + 3*{6choose2}*4*3*3^2+3*{6choose3}*3*2*3+3*{6choose4}*2$$
Yet the answer is incorret, it's a question a friend asked me to do so I'm not aware of the exact solution, from what I recall he said the answer is roughly 11,000.
combinatorics discrete-mathematics
asked Jan 14 at 10:40
MosheMoshe
396
$endgroup$
I'm required to find all possible numbers which consist of {1,2,3,4,5,6}, and include each of {1,2,3} at least once. For example: 113124 is good, but not 464612 because it's missing 3.
My approach was to first calculate all the sequences which contain 1,2,3 once. Which would be: $6times5times4times3^3$. The reasoning is that you have 6 different places to place 3, then 5 for 2, and 4 for 1. Because I'm looking now for sequences which contain each digit only once I'd fill the 3 rest spots with {4,5,6} meaning $3^3$. Now for the case where you have a digit twice it would be: $3times{6choose2}times4times3times3^2$. Using the same reasoning but this time you have $6choose2$ ways to choose 2 positions for a number, then you are left with 4 places for the other number and so on. And you multiply by 3 for each number. So you carry on up to 4 times the same number, because if you would continue for 5 you don't have enough spots to fill in the other 2 numbers you must have, so the formula is:
$$ 6*5*4*3^3 + 3*{6choose2}*4*3*3^2+3*{6choose3}*3*2*3+3*{6choose4}*2$$
Yet the answer is incorret, it's a question a friend asked me to do so I'm not aware of the exact solution, from what I recall he said the answer is roughly 11,000.
combinatorics discrete-mathematics
combinatorics discrete-mathematics
asked Jan 14 at 10:40
MosheMoshe
396
asked Jan 14 at 10:40
MosheMoshe
396
asked Jan 14 at 10:40
MosheMoshe
396
asked Jan 14 at 10:40
MosheMoshe
396
asked Jan 14 at 10:40
MosheMoshe
396
396
1
$begingroup$
Inclusion-Exclusion
$endgroup$
– Haran
Jan 14 at 10:48
add a comment |
1
$begingroup$
Inclusion-Exclusion
$endgroup$
– Haran
Jan 14 at 10:48
1
1
$begingroup$
Inclusion-Exclusion
$endgroup$
– Haran
Jan 14 at 10:48
$begingroup$
Inclusion-Exclusion
$endgroup$
– Haran
Jan 14 at 10:48
add a comment |
1 Answer
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Let $A$ denote the set of $6$-digit numbers with digits in ${1,2,3,4,5,6}$.
For $i=1,2,3,4,5,6$ let $A_i$ denote the set of elements of $A$ that do not contain $i$ as digid.
Then with inclusion/exclusion and symmetry we find:$$|A_1cup A_2cup A_3|=3|A_1|-3|A_1cap A_2|+|A_1cap A_2cap A_3|=3cdot5^6-3cdot4^6+3^6$$so that: $$|A|-|A_1cup A_2cup A_3|=6^6-3cdot5^6+3cdot4^6-3^6=11340$$
answered Jan 14 at 11:00
drhabdrhab
104k545136
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add a comment |
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1 Answer
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$begingroup$
Let $A$ denote the set of $6$-digit numbers with digits in ${1,2,3,4,5,6}$.
For $i=1,2,3,4,5,6$ let $A_i$ denote the set of elements of $A$ that do not contain $i$ as digid.
Then with inclusion/exclusion and symmetry we find:$$|A_1cup A_2cup A_3|=3|A_1|-3|A_1cap A_2|+|A_1cap A_2cap A_3|=3cdot5^6-3cdot4^6+3^6$$so that: $$|A|-|A_1cup A_2cup A_3|=6^6-3cdot5^6+3cdot4^6-3^6=11340$$
answered Jan 14 at 11:00
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Let $A$ denote the set of $6$-digit numbers with digits in ${1,2,3,4,5,6}$.
For $i=1,2,3,4,5,6$ let $A_i$ denote the set of elements of $A$ that do not contain $i$ as digid.
Then with inclusion/exclusion and symmetry we find:$$|A_1cup A_2cup A_3|=3|A_1|-3|A_1cap A_2|+|A_1cap A_2cap A_3|=3cdot5^6-3cdot4^6+3^6$$so that: $$|A|-|A_1cup A_2cup A_3|=6^6-3cdot5^6+3cdot4^6-3^6=11340$$
answered Jan 14 at 11:00
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Let $A$ denote the set of $6$-digit numbers with digits in ${1,2,3,4,5,6}$.
For $i=1,2,3,4,5,6$ let $A_i$ denote the set of elements of $A$ that do not contain $i$ as digid.
Then with inclusion/exclusion and symmetry we find:$$|A_1cup A_2cup A_3|=3|A_1|-3|A_1cap A_2|+|A_1cap A_2cap A_3|=3cdot5^6-3cdot4^6+3^6$$so that: $$|A|-|A_1cup A_2cup A_3|=6^6-3cdot5^6+3cdot4^6-3^6=11340$$
answered Jan 14 at 11:00
drhabdrhab
104k545136
$endgroup$
Let $A$ denote the set of $6$-digit numbers with digits in ${1,2,3,4,5,6}$.
For $i=1,2,3,4,5,6$ let $A_i$ denote the set of elements of $A$ that do not contain $i$ as digid.
Then with inclusion/exclusion and symmetry we find:$$|A_1cup A_2cup A_3|=3|A_1|-3|A_1cap A_2|+|A_1cap A_2cap A_3|=3cdot5^6-3cdot4^6+3^6$$so that: $$|A|-|A_1cup A_2cup A_3|=6^6-3cdot5^6+3cdot4^6-3^6=11340$$
answered Jan 14 at 11:00
drhabdrhab
104k545136
edited Jan 14 at 11:15
answered Jan 14 at 11:00
drhabdrhab
104k545136
answered Jan 14 at 11:00
drhabdrhab
104k545136
answered Jan 14 at 11:00
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
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$begingroup$
Inclusion-Exclusion
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– Haran
Jan 14 at 10:48