How many 4 digit numbers can be formed by digits 3, 1, 3, 1? [closed]
$begingroup$
This a question from my textbook and it says the answer is 6 and I fail to see how. Per my understanding if we consider each digit being used only once i.e viewing all 4 digits as distinct elements (and not two 1s and two 3s), we get the answer 4 factorial i.e 24.
combinatorics permutations combinations
asked Dec 15 '18 at 8:43
Rajdeep BiswasRajdeep Biswas
284
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closed as off-topic by Saad, amWhy, ancientmathematician, user10354138, José Carlos Santos Dec 15 '18 at 18:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, ancientmathematician, user10354138, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
This a question from my textbook and it says the answer is 6 and I fail to see how. Per my understanding if we consider each digit being used only once i.e viewing all 4 digits as distinct elements (and not two 1s and two 3s), we get the answer 4 factorial i.e 24.
combinatorics permutations combinations
asked Dec 15 '18 at 8:43
Rajdeep BiswasRajdeep Biswas
284
$endgroup$
closed as off-topic by Saad, amWhy, ancientmathematician, user10354138, José Carlos Santos Dec 15 '18 at 18:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, ancientmathematician, user10354138, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Hint : How many possible positions are there for the ones ?
$endgroup$
– Peter
Dec 15 '18 at 8:45
add a comment |
$begingroup$
This a question from my textbook and it says the answer is 6 and I fail to see how. Per my understanding if we consider each digit being used only once i.e viewing all 4 digits as distinct elements (and not two 1s and two 3s), we get the answer 4 factorial i.e 24.
combinatorics permutations combinations
asked Dec 15 '18 at 8:43
Rajdeep BiswasRajdeep Biswas
284
$endgroup$
This a question from my textbook and it says the answer is 6 and I fail to see how. Per my understanding if we consider each digit being used only once i.e viewing all 4 digits as distinct elements (and not two 1s and two 3s), we get the answer 4 factorial i.e 24.
combinatorics permutations combinations
combinatorics permutations combinations
asked Dec 15 '18 at 8:43
Rajdeep BiswasRajdeep Biswas
284
asked Dec 15 '18 at 8:43
Rajdeep BiswasRajdeep Biswas
284
asked Dec 15 '18 at 8:43
Rajdeep BiswasRajdeep Biswas
284
asked Dec 15 '18 at 8:43
Rajdeep BiswasRajdeep Biswas
284
asked Dec 15 '18 at 8:43
Rajdeep BiswasRajdeep Biswas
284
284
closed as off-topic by Saad, amWhy, ancientmathematician, user10354138, José Carlos Santos Dec 15 '18 at 18:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, ancientmathematician, user10354138, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, amWhy, ancientmathematician, user10354138, José Carlos Santos Dec 15 '18 at 18:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, ancientmathematician, user10354138, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Hint : How many possible positions are there for the ones ?
$endgroup$
– Peter
Dec 15 '18 at 8:45
add a comment |
2
$begingroup$
Hint : How many possible positions are there for the ones ?
$endgroup$
– Peter
Dec 15 '18 at 8:45
2
2
$begingroup$
Hint : How many possible positions are there for the ones ?
$endgroup$
– Peter
Dec 15 '18 at 8:45
$begingroup$
Hint : How many possible positions are there for the ones ?
$endgroup$
– Peter
Dec 15 '18 at 8:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The issue is that not every arrangement of the four digits is unique.
For example, there are multiple ways to form $1133$ - you could swap the $1$'s, the $3$'s, or both. Different rearrangements (in a sense), but the number is the same.
Thus, you need to account for that repetition. The way to handle it would be to divide by the factorials of each digit repeats. For example, we have two $3$'s, so we divide by $2!$. Since we have two $1$'s, similarly, we divide by that again.
Related example to cement this fact:
Consider the number of 8-digit numbers we can make from $9, 9, 9, 9, 7, 7, 7, 2$.
We first consider if these were unique: there are 8 digits, so the number of unique arrangements would be $8!$.
But we notice we have repeated digits. You could swap them around in the number you get, but the number would be same. You divide, then, by the number of ways you could rearrange those repeated digits (thus cutting out the repetitions).
We have $4$ $9$'s, so we will divide by $4!$. We have $3$ $7$'s, so we divide again by $3!$. We also have $1$ $2$ - we could divide by $1!$, but since $1! = 1$, we don't have to worry about that.
Thus in total, there are
$$frac{8!}{4! cdot 3!}$$
rearrangements.
answered Dec 15 '18 at 8:46
Eevee TrainerEevee Trainer
7,28221338
$endgroup$
$begingroup$
So, is the correct approach to this 4 choose 2?
$endgroup$
– Rajdeep Biswas
Dec 15 '18 at 8:49
$begingroup$
Not quite - that the same answer comes out is a mere coincidence. Notice how that doesn't generalize if, for a different example, we were told to find the number of 6-digit numbers for $4, 3, 3, 2, 2, 2$. What the correct/intended method is, I wouldn't know since I haven't taken a combinatorics course. My method would be just to notice that, for each individual number, you have three superfluous numbers: you can swap the $1$'s for one of them, swap the $3$'s for another, and swap both pairs for a third.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
So then you notice that each "unique" number has three duplicates as well, i.e. $3/4$ of the permutations (were the digits unique) are pointless. Or more properly, only $1/4$ of the numbers you would get by permuting them as if they were unique are actually unique. Thus, the number of unique numbers would be $$frac{1}{4} cdot 4!$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
Granted that's just how I personally would handle it. How a textbook/classroom setting would expect you to handle it, I don't know, owing to not having been in those environments.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:55
$begingroup$
Looking at the other answer, the proper way would be to divide by the number of repeated digits. For example, for our 6-digit problem earlier, the answer would be $$frac{6!}{2! cdot 3!}$$ Or for the number of 10-digit numbers made by $1, 2, 2, 3, 3, 3, 4, 4, 4, 4$: $$frac{10!}{2! cdot 3! cdot 4!}$$ because we have two $2$'s, three $3$'s, and four $4$'s.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:58
|
show 3 more comments
$begingroup$
If the 4 digits are distinct then you'll have $4!=24$ ways to to form the number.
But since there is 2 $1$'s, we divide by $2!$
And since there is 2 $3$'s, we divide another time by $2!$
So the number of ways to form the number is $frac{24}{2!.2!} =6$
Note that we divide by $2!$ to avoid the repetition of the same number twice i.e. 1133 will stay the same if we swap the $1$'s. So by dividing by $2!$ we will be counting the number $1133$ once and not twice.
For example when you have 3 $1$'s you divide the total number of ways by $3!$ to avoid the repetition of the same number just by swapping the $1$'s
answered Dec 15 '18 at 8:55
Fareed AFFareed AF
57112
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The issue is that not every arrangement of the four digits is unique.
For example, there are multiple ways to form $1133$ - you could swap the $1$'s, the $3$'s, or both. Different rearrangements (in a sense), but the number is the same.
Thus, you need to account for that repetition. The way to handle it would be to divide by the factorials of each digit repeats. For example, we have two $3$'s, so we divide by $2!$. Since we have two $1$'s, similarly, we divide by that again.
Related example to cement this fact:
Consider the number of 8-digit numbers we can make from $9, 9, 9, 9, 7, 7, 7, 2$.
We first consider if these were unique: there are 8 digits, so the number of unique arrangements would be $8!$.
But we notice we have repeated digits. You could swap them around in the number you get, but the number would be same. You divide, then, by the number of ways you could rearrange those repeated digits (thus cutting out the repetitions).
We have $4$ $9$'s, so we will divide by $4!$. We have $3$ $7$'s, so we divide again by $3!$. We also have $1$ $2$ - we could divide by $1!$, but since $1! = 1$, we don't have to worry about that.
Thus in total, there are
$$frac{8!}{4! cdot 3!}$$
rearrangements.
answered Dec 15 '18 at 8:46
Eevee TrainerEevee Trainer
7,28221338
$endgroup$
$begingroup$
So, is the correct approach to this 4 choose 2?
$endgroup$
– Rajdeep Biswas
Dec 15 '18 at 8:49
$begingroup$
Not quite - that the same answer comes out is a mere coincidence. Notice how that doesn't generalize if, for a different example, we were told to find the number of 6-digit numbers for $4, 3, 3, 2, 2, 2$. What the correct/intended method is, I wouldn't know since I haven't taken a combinatorics course. My method would be just to notice that, for each individual number, you have three superfluous numbers: you can swap the $1$'s for one of them, swap the $3$'s for another, and swap both pairs for a third.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
So then you notice that each "unique" number has three duplicates as well, i.e. $3/4$ of the permutations (were the digits unique) are pointless. Or more properly, only $1/4$ of the numbers you would get by permuting them as if they were unique are actually unique. Thus, the number of unique numbers would be $$frac{1}{4} cdot 4!$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
Granted that's just how I personally would handle it. How a textbook/classroom setting would expect you to handle it, I don't know, owing to not having been in those environments.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:55
$begingroup$
Looking at the other answer, the proper way would be to divide by the number of repeated digits. For example, for our 6-digit problem earlier, the answer would be $$frac{6!}{2! cdot 3!}$$ Or for the number of 10-digit numbers made by $1, 2, 2, 3, 3, 3, 4, 4, 4, 4$: $$frac{10!}{2! cdot 3! cdot 4!}$$ because we have two $2$'s, three $3$'s, and four $4$'s.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:58
|
show 3 more comments
$begingroup$
The issue is that not every arrangement of the four digits is unique.
For example, there are multiple ways to form $1133$ - you could swap the $1$'s, the $3$'s, or both. Different rearrangements (in a sense), but the number is the same.
Thus, you need to account for that repetition. The way to handle it would be to divide by the factorials of each digit repeats. For example, we have two $3$'s, so we divide by $2!$. Since we have two $1$'s, similarly, we divide by that again.
Related example to cement this fact:
Consider the number of 8-digit numbers we can make from $9, 9, 9, 9, 7, 7, 7, 2$.
We first consider if these were unique: there are 8 digits, so the number of unique arrangements would be $8!$.
But we notice we have repeated digits. You could swap them around in the number you get, but the number would be same. You divide, then, by the number of ways you could rearrange those repeated digits (thus cutting out the repetitions).
We have $4$ $9$'s, so we will divide by $4!$. We have $3$ $7$'s, so we divide again by $3!$. We also have $1$ $2$ - we could divide by $1!$, but since $1! = 1$, we don't have to worry about that.
Thus in total, there are
$$frac{8!}{4! cdot 3!}$$
rearrangements.
answered Dec 15 '18 at 8:46
Eevee TrainerEevee Trainer
7,28221338
$endgroup$
$begingroup$
So, is the correct approach to this 4 choose 2?
$endgroup$
– Rajdeep Biswas
Dec 15 '18 at 8:49
$begingroup$
Not quite - that the same answer comes out is a mere coincidence. Notice how that doesn't generalize if, for a different example, we were told to find the number of 6-digit numbers for $4, 3, 3, 2, 2, 2$. What the correct/intended method is, I wouldn't know since I haven't taken a combinatorics course. My method would be just to notice that, for each individual number, you have three superfluous numbers: you can swap the $1$'s for one of them, swap the $3$'s for another, and swap both pairs for a third.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
So then you notice that each "unique" number has three duplicates as well, i.e. $3/4$ of the permutations (were the digits unique) are pointless. Or more properly, only $1/4$ of the numbers you would get by permuting them as if they were unique are actually unique. Thus, the number of unique numbers would be $$frac{1}{4} cdot 4!$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
Granted that's just how I personally would handle it. How a textbook/classroom setting would expect you to handle it, I don't know, owing to not having been in those environments.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:55
$begingroup$
Looking at the other answer, the proper way would be to divide by the number of repeated digits. For example, for our 6-digit problem earlier, the answer would be $$frac{6!}{2! cdot 3!}$$ Or for the number of 10-digit numbers made by $1, 2, 2, 3, 3, 3, 4, 4, 4, 4$: $$frac{10!}{2! cdot 3! cdot 4!}$$ because we have two $2$'s, three $3$'s, and four $4$'s.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:58
|
show 3 more comments
$begingroup$
The issue is that not every arrangement of the four digits is unique.
For example, there are multiple ways to form $1133$ - you could swap the $1$'s, the $3$'s, or both. Different rearrangements (in a sense), but the number is the same.
Thus, you need to account for that repetition. The way to handle it would be to divide by the factorials of each digit repeats. For example, we have two $3$'s, so we divide by $2!$. Since we have two $1$'s, similarly, we divide by that again.
Related example to cement this fact:
Consider the number of 8-digit numbers we can make from $9, 9, 9, 9, 7, 7, 7, 2$.
We first consider if these were unique: there are 8 digits, so the number of unique arrangements would be $8!$.
But we notice we have repeated digits. You could swap them around in the number you get, but the number would be same. You divide, then, by the number of ways you could rearrange those repeated digits (thus cutting out the repetitions).
We have $4$ $9$'s, so we will divide by $4!$. We have $3$ $7$'s, so we divide again by $3!$. We also have $1$ $2$ - we could divide by $1!$, but since $1! = 1$, we don't have to worry about that.
Thus in total, there are
$$frac{8!}{4! cdot 3!}$$
rearrangements.
answered Dec 15 '18 at 8:46
Eevee TrainerEevee Trainer
7,28221338
$endgroup$
The issue is that not every arrangement of the four digits is unique.
For example, there are multiple ways to form $1133$ - you could swap the $1$'s, the $3$'s, or both. Different rearrangements (in a sense), but the number is the same.
Thus, you need to account for that repetition. The way to handle it would be to divide by the factorials of each digit repeats. For example, we have two $3$'s, so we divide by $2!$. Since we have two $1$'s, similarly, we divide by that again.
Related example to cement this fact:
Consider the number of 8-digit numbers we can make from $9, 9, 9, 9, 7, 7, 7, 2$.
We first consider if these were unique: there are 8 digits, so the number of unique arrangements would be $8!$.
But we notice we have repeated digits. You could swap them around in the number you get, but the number would be same. You divide, then, by the number of ways you could rearrange those repeated digits (thus cutting out the repetitions).
We have $4$ $9$'s, so we will divide by $4!$. We have $3$ $7$'s, so we divide again by $3!$. We also have $1$ $2$ - we could divide by $1!$, but since $1! = 1$, we don't have to worry about that.
Thus in total, there are
$$frac{8!}{4! cdot 3!}$$
rearrangements.
answered Dec 15 '18 at 8:46
Eevee TrainerEevee Trainer
7,28221338
edited Dec 15 '18 at 9:05
answered Dec 15 '18 at 8:46
Eevee TrainerEevee Trainer
7,28221338
answered Dec 15 '18 at 8:46
Eevee TrainerEevee Trainer
7,28221338
answered Dec 15 '18 at 8:46
Eevee TrainerEevee Trainer
7,28221338
7,28221338
$begingroup$
So, is the correct approach to this 4 choose 2?
$endgroup$
– Rajdeep Biswas
Dec 15 '18 at 8:49
$begingroup$
Not quite - that the same answer comes out is a mere coincidence. Notice how that doesn't generalize if, for a different example, we were told to find the number of 6-digit numbers for $4, 3, 3, 2, 2, 2$. What the correct/intended method is, I wouldn't know since I haven't taken a combinatorics course. My method would be just to notice that, for each individual number, you have three superfluous numbers: you can swap the $1$'s for one of them, swap the $3$'s for another, and swap both pairs for a third.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
So then you notice that each "unique" number has three duplicates as well, i.e. $3/4$ of the permutations (were the digits unique) are pointless. Or more properly, only $1/4$ of the numbers you would get by permuting them as if they were unique are actually unique. Thus, the number of unique numbers would be $$frac{1}{4} cdot 4!$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
Granted that's just how I personally would handle it. How a textbook/classroom setting would expect you to handle it, I don't know, owing to not having been in those environments.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:55
$begingroup$
Looking at the other answer, the proper way would be to divide by the number of repeated digits. For example, for our 6-digit problem earlier, the answer would be $$frac{6!}{2! cdot 3!}$$ Or for the number of 10-digit numbers made by $1, 2, 2, 3, 3, 3, 4, 4, 4, 4$: $$frac{10!}{2! cdot 3! cdot 4!}$$ because we have two $2$'s, three $3$'s, and four $4$'s.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:58
|
show 3 more comments
$begingroup$
So, is the correct approach to this 4 choose 2?
$endgroup$
– Rajdeep Biswas
Dec 15 '18 at 8:49
$begingroup$
Not quite - that the same answer comes out is a mere coincidence. Notice how that doesn't generalize if, for a different example, we were told to find the number of 6-digit numbers for $4, 3, 3, 2, 2, 2$. What the correct/intended method is, I wouldn't know since I haven't taken a combinatorics course. My method would be just to notice that, for each individual number, you have three superfluous numbers: you can swap the $1$'s for one of them, swap the $3$'s for another, and swap both pairs for a third.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
So then you notice that each "unique" number has three duplicates as well, i.e. $3/4$ of the permutations (were the digits unique) are pointless. Or more properly, only $1/4$ of the numbers you would get by permuting them as if they were unique are actually unique. Thus, the number of unique numbers would be $$frac{1}{4} cdot 4!$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
Granted that's just how I personally would handle it. How a textbook/classroom setting would expect you to handle it, I don't know, owing to not having been in those environments.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:55
$begingroup$
Looking at the other answer, the proper way would be to divide by the number of repeated digits. For example, for our 6-digit problem earlier, the answer would be $$frac{6!}{2! cdot 3!}$$ Or for the number of 10-digit numbers made by $1, 2, 2, 3, 3, 3, 4, 4, 4, 4$: $$frac{10!}{2! cdot 3! cdot 4!}$$ because we have two $2$'s, three $3$'s, and four $4$'s.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:58
$begingroup$
So, is the correct approach to this 4 choose 2?
$endgroup$
– Rajdeep Biswas
Dec 15 '18 at 8:49
$begingroup$
So, is the correct approach to this 4 choose 2?
$endgroup$
– Rajdeep Biswas
Dec 15 '18 at 8:49
$begingroup$
Not quite - that the same answer comes out is a mere coincidence. Notice how that doesn't generalize if, for a different example, we were told to find the number of 6-digit numbers for $4, 3, 3, 2, 2, 2$. What the correct/intended method is, I wouldn't know since I haven't taken a combinatorics course. My method would be just to notice that, for each individual number, you have three superfluous numbers: you can swap the $1$'s for one of them, swap the $3$'s for another, and swap both pairs for a third.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
Not quite - that the same answer comes out is a mere coincidence. Notice how that doesn't generalize if, for a different example, we were told to find the number of 6-digit numbers for $4, 3, 3, 2, 2, 2$. What the correct/intended method is, I wouldn't know since I haven't taken a combinatorics course. My method would be just to notice that, for each individual number, you have three superfluous numbers: you can swap the $1$'s for one of them, swap the $3$'s for another, and swap both pairs for a third.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
So then you notice that each "unique" number has three duplicates as well, i.e. $3/4$ of the permutations (were the digits unique) are pointless. Or more properly, only $1/4$ of the numbers you would get by permuting them as if they were unique are actually unique. Thus, the number of unique numbers would be $$frac{1}{4} cdot 4!$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
So then you notice that each "unique" number has three duplicates as well, i.e. $3/4$ of the permutations (were the digits unique) are pointless. Or more properly, only $1/4$ of the numbers you would get by permuting them as if they were unique are actually unique. Thus, the number of unique numbers would be $$frac{1}{4} cdot 4!$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:54
$begingroup$
Granted that's just how I personally would handle it. How a textbook/classroom setting would expect you to handle it, I don't know, owing to not having been in those environments.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:55
$begingroup$
Granted that's just how I personally would handle it. How a textbook/classroom setting would expect you to handle it, I don't know, owing to not having been in those environments.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:55
$begingroup$
Looking at the other answer, the proper way would be to divide by the number of repeated digits. For example, for our 6-digit problem earlier, the answer would be $$frac{6!}{2! cdot 3!}$$ Or for the number of 10-digit numbers made by $1, 2, 2, 3, 3, 3, 4, 4, 4, 4$: $$frac{10!}{2! cdot 3! cdot 4!}$$ because we have two $2$'s, three $3$'s, and four $4$'s.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:58
$begingroup$
Looking at the other answer, the proper way would be to divide by the number of repeated digits. For example, for our 6-digit problem earlier, the answer would be $$frac{6!}{2! cdot 3!}$$ Or for the number of 10-digit numbers made by $1, 2, 2, 3, 3, 3, 4, 4, 4, 4$: $$frac{10!}{2! cdot 3! cdot 4!}$$ because we have two $2$'s, three $3$'s, and four $4$'s.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 8:58
|
show 3 more comments
$begingroup$
If the 4 digits are distinct then you'll have $4!=24$ ways to to form the number.
But since there is 2 $1$'s, we divide by $2!$
And since there is 2 $3$'s, we divide another time by $2!$
So the number of ways to form the number is $frac{24}{2!.2!} =6$
Note that we divide by $2!$ to avoid the repetition of the same number twice i.e. 1133 will stay the same if we swap the $1$'s. So by dividing by $2!$ we will be counting the number $1133$ once and not twice.
For example when you have 3 $1$'s you divide the total number of ways by $3!$ to avoid the repetition of the same number just by swapping the $1$'s
answered Dec 15 '18 at 8:55
Fareed AFFareed AF
57112
$endgroup$
add a comment |
$begingroup$
If the 4 digits are distinct then you'll have $4!=24$ ways to to form the number.
But since there is 2 $1$'s, we divide by $2!$
And since there is 2 $3$'s, we divide another time by $2!$
So the number of ways to form the number is $frac{24}{2!.2!} =6$
Note that we divide by $2!$ to avoid the repetition of the same number twice i.e. 1133 will stay the same if we swap the $1$'s. So by dividing by $2!$ we will be counting the number $1133$ once and not twice.
For example when you have 3 $1$'s you divide the total number of ways by $3!$ to avoid the repetition of the same number just by swapping the $1$'s
answered Dec 15 '18 at 8:55
Fareed AFFareed AF
57112
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$begingroup$
If the 4 digits are distinct then you'll have $4!=24$ ways to to form the number.
But since there is 2 $1$'s, we divide by $2!$
And since there is 2 $3$'s, we divide another time by $2!$
So the number of ways to form the number is $frac{24}{2!.2!} =6$
Note that we divide by $2!$ to avoid the repetition of the same number twice i.e. 1133 will stay the same if we swap the $1$'s. So by dividing by $2!$ we will be counting the number $1133$ once and not twice.
For example when you have 3 $1$'s you divide the total number of ways by $3!$ to avoid the repetition of the same number just by swapping the $1$'s
answered Dec 15 '18 at 8:55
Fareed AFFareed AF
57112
$endgroup$
If the 4 digits are distinct then you'll have $4!=24$ ways to to form the number.
But since there is 2 $1$'s, we divide by $2!$
And since there is 2 $3$'s, we divide another time by $2!$
So the number of ways to form the number is $frac{24}{2!.2!} =6$
Note that we divide by $2!$ to avoid the repetition of the same number twice i.e. 1133 will stay the same if we swap the $1$'s. So by dividing by $2!$ we will be counting the number $1133$ once and not twice.
For example when you have 3 $1$'s you divide the total number of ways by $3!$ to avoid the repetition of the same number just by swapping the $1$'s
answered Dec 15 '18 at 8:55
Fareed AFFareed AF
57112
edited Dec 15 '18 at 9:00
answered Dec 15 '18 at 8:55
Fareed AFFareed AF
57112
answered Dec 15 '18 at 8:55
Fareed AFFareed AF
57112
answered Dec 15 '18 at 8:55
Fareed AFFareed AF
57112
57112
add a comment |
add a comment |
2
$begingroup$
Hint : How many possible positions are there for the ones ?
$endgroup$
– Peter
Dec 15 '18 at 8:45