Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?
Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?
one of the digits which can be formed is $4444$
$4$ digit numbers greater than $3000$, which consists of only $2's$ and $4's$ are $4224$, $4242$, $4244$, $4422$, $4424$, $4442$
is there a well defined technique to solve this question.
combinatorics
asked Aug 26 '12 at 3:40
HOLYBIBLETHE
1,45221837
add a comment |
Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?
one of the digits which can be formed is $4444$
$4$ digit numbers greater than $3000$, which consists of only $2's$ and $4's$ are $4224$, $4242$, $4244$, $4422$, $4424$, $4442$
is there a well defined technique to solve this question.
combinatorics
asked Aug 26 '12 at 3:40
HOLYBIBLETHE
1,45221837
add a comment |
Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?
one of the digits which can be formed is $4444$
$4$ digit numbers greater than $3000$, which consists of only $2's$ and $4's$ are $4224$, $4242$, $4244$, $4422$, $4424$, $4442$
is there a well defined technique to solve this question.
combinatorics
asked Aug 26 '12 at 3:40
HOLYBIBLETHE
1,45221837
Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?
one of the digits which can be formed is $4444$
$4$ digit numbers greater than $3000$, which consists of only $2's$ and $4's$ are $4224$, $4242$, $4244$, $4422$, $4424$, $4442$
is there a well defined technique to solve this question.
combinatorics
combinatorics
asked Aug 26 '12 at 3:40
HOLYBIBLETHE
1,45221837
asked Aug 26 '12 at 3:40
HOLYBIBLETHE
1,45221837
edited Aug 26 '12 at 3:54
asked Aug 26 '12 at 3:40
HOLYBIBLETHE
1,45221837
asked Aug 26 '12 at 3:40
HOLYBIBLETHE
1,45221837
asked Aug 26 '12 at 3:40
HOLYBIBLETHE
1,45221837
1,45221837
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.
First place can have either $3$ or $4$. So we have two choices. Lets analyse it.
If first place is $3$ - so we have to choose 3 digits from $(2,2,3,4,4,4,4)$. So any place can have either 2 or 3 or 4, but $(222,333,332,323,233,334,343,433)$ is not possible, because we have just one $3$'s and two $2$'s
So total = $3 times 3times 3 - 8 = 19$
If first place is $4$ - so now we have to choose 3 digits from $(2,2,3,3,4,4,4)$. So any place can have either $2$ or $3$ or $4$, but $(222,333)$ is not possible, because we have just two $3$'s and two $2$'s
So total = $3 times 3times 3 - 2 = 25$
So, total choices = $19+25 = 44$, which is your answer.
edited Jun 5 '13 at 7:00
learner
3,37032167
answered Aug 26 '12 at 7:16
Rahul
399310
There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
– ladaghini
Aug 26 '12 at 7:21
@ladaghini: please see the edited answer. I previously applied a wrong approach.
– Rahul
Aug 26 '12 at 7:54
But 44 is not an option. The correct answer is given as 51. I wonder how.
– Ramit
Dec 30 '16 at 4:53
add a comment |
It is a matter of detailed counting. You have to start with a $3$ or $4$ to be greater than $3000$. If you start with a $4$, you have three choices for each other space, except you can't have $222$ or $333$, so there are $25$. If you start with a $3$ it is harder as you have a different number of each digit left, but it is the same idea.
answered Aug 26 '12 at 4:09
Ross Millikan
291k23196370
Can you provide your final answer please?
– Ramit
Dec 30 '16 at 4:54
@Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
– Ross Millikan
Dec 30 '16 at 5:29
add a comment |
I will use the fundamental principle of counting to solve this question.
Given the set of numbers, $D = {2, 2, 3, 3, 3, 4, 4, 4, 4}$.
Here, count(2's) = 2, count(3's) = 3, count(4's) = 4.
We are required to construct 4-digit numbers greater than 3000. For easy visualisation, we will use dashes on paper, _ _ _ _.
Case 1: Thousandth's place is 3.
Set, $D' = {2, 2, 3, 3, 4, 4, 4, 4}$.
Rest of the 3 places can be filled in $3 times 3 times 3$ ways provided we have 3 counts of the three unique digits, for filling each of the 3 places. But count(2's) = 2, count(3's) = 2 in the new set. Deficiency for each digit is 1. (Impossible numbers - 3222, 3333).
Therefore, ways to fill = $3 times 3 times 3 - (1+1) = 25$.
Case 2: Thousandth's place is 4.
Set, $D' = {2, 2, 3, 3, 3, 4, 4, 4}$.
Similarly, the deficiency is of only digit 2, which is 1 count. (Impossible number - 4222)
Therefore ways to fill = $3 times 3 times 3 - 1 = 26$.
Summing each case up, we get $26 + 25 = 51$.
Hope this helps.
answered Oct 26 at 4:19
Manoj Baishya
12
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.
First place can have either $3$ or $4$. So we have two choices. Lets analyse it.
If first place is $3$ - so we have to choose 3 digits from $(2,2,3,4,4,4,4)$. So any place can have either 2 or 3 or 4, but $(222,333,332,323,233,334,343,433)$ is not possible, because we have just one $3$'s and two $2$'s
So total = $3 times 3times 3 - 8 = 19$
If first place is $4$ - so now we have to choose 3 digits from $(2,2,3,3,4,4,4)$. So any place can have either $2$ or $3$ or $4$, but $(222,333)$ is not possible, because we have just two $3$'s and two $2$'s
So total = $3 times 3times 3 - 2 = 25$
So, total choices = $19+25 = 44$, which is your answer.
edited Jun 5 '13 at 7:00
learner
3,37032167
answered Aug 26 '12 at 7:16
Rahul
399310
There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
– ladaghini
Aug 26 '12 at 7:21
@ladaghini: please see the edited answer. I previously applied a wrong approach.
– Rahul
Aug 26 '12 at 7:54
But 44 is not an option. The correct answer is given as 51. I wonder how.
– Ramit
Dec 30 '16 at 4:53
add a comment |
You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.
First place can have either $3$ or $4$. So we have two choices. Lets analyse it.
If first place is $3$ - so we have to choose 3 digits from $(2,2,3,4,4,4,4)$. So any place can have either 2 or 3 or 4, but $(222,333,332,323,233,334,343,433)$ is not possible, because we have just one $3$'s and two $2$'s
So total = $3 times 3times 3 - 8 = 19$
If first place is $4$ - so now we have to choose 3 digits from $(2,2,3,3,4,4,4)$. So any place can have either $2$ or $3$ or $4$, but $(222,333)$ is not possible, because we have just two $3$'s and two $2$'s
So total = $3 times 3times 3 - 2 = 25$
So, total choices = $19+25 = 44$, which is your answer.
edited Jun 5 '13 at 7:00
learner
3,37032167
answered Aug 26 '12 at 7:16
Rahul
399310
There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
– ladaghini
Aug 26 '12 at 7:21
@ladaghini: please see the edited answer. I previously applied a wrong approach.
– Rahul
Aug 26 '12 at 7:54
But 44 is not an option. The correct answer is given as 51. I wonder how.
– Ramit
Dec 30 '16 at 4:53
add a comment |
You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.
First place can have either $3$ or $4$. So we have two choices. Lets analyse it.
If first place is $3$ - so we have to choose 3 digits from $(2,2,3,4,4,4,4)$. So any place can have either 2 or 3 or 4, but $(222,333,332,323,233,334,343,433)$ is not possible, because we have just one $3$'s and two $2$'s
So total = $3 times 3times 3 - 8 = 19$
If first place is $4$ - so now we have to choose 3 digits from $(2,2,3,3,4,4,4)$. So any place can have either $2$ or $3$ or $4$, but $(222,333)$ is not possible, because we have just two $3$'s and two $2$'s
So total = $3 times 3times 3 - 2 = 25$
So, total choices = $19+25 = 44$, which is your answer.
edited Jun 5 '13 at 7:00
learner
3,37032167
answered Aug 26 '12 at 7:16
Rahul
399310
You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.
First place can have either $3$ or $4$. So we have two choices. Lets analyse it.
If first place is $3$ - so we have to choose 3 digits from $(2,2,3,4,4,4,4)$. So any place can have either 2 or 3 or 4, but $(222,333,332,323,233,334,343,433)$ is not possible, because we have just one $3$'s and two $2$'s
So total = $3 times 3times 3 - 8 = 19$
If first place is $4$ - so now we have to choose 3 digits from $(2,2,3,3,4,4,4)$. So any place can have either $2$ or $3$ or $4$, but $(222,333)$ is not possible, because we have just two $3$'s and two $2$'s
So total = $3 times 3times 3 - 2 = 25$
So, total choices = $19+25 = 44$, which is your answer.
edited Jun 5 '13 at 7:00
learner
3,37032167
answered Aug 26 '12 at 7:16
Rahul
399310
edited Jun 5 '13 at 7:00
learner
3,37032167
edited Jun 5 '13 at 7:00
learner
3,37032167
edited Jun 5 '13 at 7:00
learner
3,37032167
3,37032167
answered Aug 26 '12 at 7:16
Rahul
399310
answered Aug 26 '12 at 7:16
Rahul
399310
answered Aug 26 '12 at 7:16
Rahul
399310
399310
There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
– ladaghini
Aug 26 '12 at 7:21
@ladaghini: please see the edited answer. I previously applied a wrong approach.
– Rahul
Aug 26 '12 at 7:54
But 44 is not an option. The correct answer is given as 51. I wonder how.
– Ramit
Dec 30 '16 at 4:53
add a comment |
There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
– ladaghini
Aug 26 '12 at 7:21
@ladaghini: please see the edited answer. I previously applied a wrong approach.
– Rahul
Aug 26 '12 at 7:54
But 44 is not an option. The correct answer is given as 51. I wonder how.
– Ramit
Dec 30 '16 at 4:53
There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
– ladaghini
Aug 26 '12 at 7:21
There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
– ladaghini
Aug 26 '12 at 7:21
@ladaghini: please see the edited answer. I previously applied a wrong approach.
– Rahul
Aug 26 '12 at 7:54
@ladaghini: please see the edited answer. I previously applied a wrong approach.
– Rahul
Aug 26 '12 at 7:54
But 44 is not an option. The correct answer is given as 51. I wonder how.
– Ramit
Dec 30 '16 at 4:53
But 44 is not an option. The correct answer is given as 51. I wonder how.
– Ramit
Dec 30 '16 at 4:53
add a comment |
It is a matter of detailed counting. You have to start with a $3$ or $4$ to be greater than $3000$. If you start with a $4$, you have three choices for each other space, except you can't have $222$ or $333$, so there are $25$. If you start with a $3$ it is harder as you have a different number of each digit left, but it is the same idea.
answered Aug 26 '12 at 4:09
Ross Millikan
291k23196370
Can you provide your final answer please?
– Ramit
Dec 30 '16 at 4:54
@Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
– Ross Millikan
Dec 30 '16 at 5:29
add a comment |
It is a matter of detailed counting. You have to start with a $3$ or $4$ to be greater than $3000$. If you start with a $4$, you have three choices for each other space, except you can't have $222$ or $333$, so there are $25$. If you start with a $3$ it is harder as you have a different number of each digit left, but it is the same idea.
answered Aug 26 '12 at 4:09
Ross Millikan
291k23196370
Can you provide your final answer please?
– Ramit
Dec 30 '16 at 4:54
@Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
– Ross Millikan
Dec 30 '16 at 5:29
add a comment |
It is a matter of detailed counting. You have to start with a $3$ or $4$ to be greater than $3000$. If you start with a $4$, you have three choices for each other space, except you can't have $222$ or $333$, so there are $25$. If you start with a $3$ it is harder as you have a different number of each digit left, but it is the same idea.
answered Aug 26 '12 at 4:09
Ross Millikan
291k23196370
It is a matter of detailed counting. You have to start with a $3$ or $4$ to be greater than $3000$. If you start with a $4$, you have three choices for each other space, except you can't have $222$ or $333$, so there are $25$. If you start with a $3$ it is harder as you have a different number of each digit left, but it is the same idea.
answered Aug 26 '12 at 4:09
Ross Millikan
291k23196370
answered Aug 26 '12 at 4:09
Ross Millikan
291k23196370
answered Aug 26 '12 at 4:09
Ross Millikan
291k23196370
answered Aug 26 '12 at 4:09
Ross Millikan
291k23196370
291k23196370
Can you provide your final answer please?
– Ramit
Dec 30 '16 at 4:54
@Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
– Ross Millikan
Dec 30 '16 at 5:29
add a comment |
Can you provide your final answer please?
– Ramit
Dec 30 '16 at 4:54
@Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
– Ross Millikan
Dec 30 '16 at 5:29
Can you provide your final answer please?
– Ramit
Dec 30 '16 at 4:54
Can you provide your final answer please?
– Ramit
Dec 30 '16 at 4:54
@Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
– Ross Millikan
Dec 30 '16 at 5:29
@Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
– Ross Millikan
Dec 30 '16 at 5:29
add a comment |
I will use the fundamental principle of counting to solve this question.
Given the set of numbers, $D = {2, 2, 3, 3, 3, 4, 4, 4, 4}$.
Here, count(2's) = 2, count(3's) = 3, count(4's) = 4.
We are required to construct 4-digit numbers greater than 3000. For easy visualisation, we will use dashes on paper, _ _ _ _.
Case 1: Thousandth's place is 3.
Set, $D' = {2, 2, 3, 3, 4, 4, 4, 4}$.
Rest of the 3 places can be filled in $3 times 3 times 3$ ways provided we have 3 counts of the three unique digits, for filling each of the 3 places. But count(2's) = 2, count(3's) = 2 in the new set. Deficiency for each digit is 1. (Impossible numbers - 3222, 3333).
Therefore, ways to fill = $3 times 3 times 3 - (1+1) = 25$.
Case 2: Thousandth's place is 4.
Set, $D' = {2, 2, 3, 3, 3, 4, 4, 4}$.
Similarly, the deficiency is of only digit 2, which is 1 count. (Impossible number - 4222)
Therefore ways to fill = $3 times 3 times 3 - 1 = 26$.
Summing each case up, we get $26 + 25 = 51$.
Hope this helps.
answered Oct 26 at 4:19
Manoj Baishya
12
add a comment |
I will use the fundamental principle of counting to solve this question.
Given the set of numbers, $D = {2, 2, 3, 3, 3, 4, 4, 4, 4}$.
Here, count(2's) = 2, count(3's) = 3, count(4's) = 4.
We are required to construct 4-digit numbers greater than 3000. For easy visualisation, we will use dashes on paper, _ _ _ _.
Case 1: Thousandth's place is 3.
Set, $D' = {2, 2, 3, 3, 4, 4, 4, 4}$.
Rest of the 3 places can be filled in $3 times 3 times 3$ ways provided we have 3 counts of the three unique digits, for filling each of the 3 places. But count(2's) = 2, count(3's) = 2 in the new set. Deficiency for each digit is 1. (Impossible numbers - 3222, 3333).
Therefore, ways to fill = $3 times 3 times 3 - (1+1) = 25$.
Case 2: Thousandth's place is 4.
Set, $D' = {2, 2, 3, 3, 3, 4, 4, 4}$.
Similarly, the deficiency is of only digit 2, which is 1 count. (Impossible number - 4222)
Therefore ways to fill = $3 times 3 times 3 - 1 = 26$.
Summing each case up, we get $26 + 25 = 51$.
Hope this helps.
answered Oct 26 at 4:19
Manoj Baishya
12
add a comment |
I will use the fundamental principle of counting to solve this question.
Given the set of numbers, $D = {2, 2, 3, 3, 3, 4, 4, 4, 4}$.
Here, count(2's) = 2, count(3's) = 3, count(4's) = 4.
We are required to construct 4-digit numbers greater than 3000. For easy visualisation, we will use dashes on paper, _ _ _ _.
Case 1: Thousandth's place is 3.
Set, $D' = {2, 2, 3, 3, 4, 4, 4, 4}$.
Rest of the 3 places can be filled in $3 times 3 times 3$ ways provided we have 3 counts of the three unique digits, for filling each of the 3 places. But count(2's) = 2, count(3's) = 2 in the new set. Deficiency for each digit is 1. (Impossible numbers - 3222, 3333).
Therefore, ways to fill = $3 times 3 times 3 - (1+1) = 25$.
Case 2: Thousandth's place is 4.
Set, $D' = {2, 2, 3, 3, 3, 4, 4, 4}$.
Similarly, the deficiency is of only digit 2, which is 1 count. (Impossible number - 4222)
Therefore ways to fill = $3 times 3 times 3 - 1 = 26$.
Summing each case up, we get $26 + 25 = 51$.
Hope this helps.
answered Oct 26 at 4:19
Manoj Baishya
12
I will use the fundamental principle of counting to solve this question.
Given the set of numbers, $D = {2, 2, 3, 3, 3, 4, 4, 4, 4}$.
Here, count(2's) = 2, count(3's) = 3, count(4's) = 4.
We are required to construct 4-digit numbers greater than 3000. For easy visualisation, we will use dashes on paper, _ _ _ _.
Case 1: Thousandth's place is 3.
Set, $D' = {2, 2, 3, 3, 4, 4, 4, 4}$.
Rest of the 3 places can be filled in $3 times 3 times 3$ ways provided we have 3 counts of the three unique digits, for filling each of the 3 places. But count(2's) = 2, count(3's) = 2 in the new set. Deficiency for each digit is 1. (Impossible numbers - 3222, 3333).
Therefore, ways to fill = $3 times 3 times 3 - (1+1) = 25$.
Case 2: Thousandth's place is 4.
Set, $D' = {2, 2, 3, 3, 3, 4, 4, 4}$.
Similarly, the deficiency is of only digit 2, which is 1 count. (Impossible number - 4222)
Therefore ways to fill = $3 times 3 times 3 - 1 = 26$.
Summing each case up, we get $26 + 25 = 51$.
Hope this helps.
answered Oct 26 at 4:19
Manoj Baishya
12
edited Dec 7 at 13:04
answered Oct 26 at 4:19
Manoj Baishya
12
answered Oct 26 at 4:19
Manoj Baishya
12
answered Oct 26 at 4:19
Manoj Baishya
12
12
add a comment |
add a comment |
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