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Using the numbers 1, 2, 3, 4 and 5, how many 3-digit combinations can you make? Repetitions allowed. [closed]

Calculating the probabilities of different lengths of repetitions of numbers of length 6You bought six numbers at your local hardware store. The numbers are 0, 1, 2, 3, 4, 5.How many lists of 100 numbers (1 to 10 only) add to 700?How many 5 digit numbers can be formed out of 1,2,3…,9 where a digit can repeat at most twice?In how many ways can five-digit numbers be formed by using digits $0,2,4,6,8$ such that the numbers are divisible by $8$?How many possible numbers do I have?How can we count combinations with repetition (or permutations) using inequality symbols between each number?Arrangements of 3 baskets, 2 misses through Combinations or Permutations?In how many permutations of $1,2,3…100$ will the 25th number be the minimum of the first 25 numbers, and likewise for the 50th of the first 50?How many $4$-digit lock combinations are possible if each digit in the code may appear at most twice?

0

$begingroup$

The question is:

Using the numbers $1, 2, 3, 4,$ and $5$, how many $3$-digit combinations can you make? Repetitions are allowed. Thanks!

Edit:

I just watched the video Total 3 Digit Numbers If Repetition allowed & Not allowed - Permutations & Combinations Problems. That answered my question.

But now I'd like to add some limitations, i. e.: I don't want more than $2$ repeats per $3$ combo. So, is it correct to say $5$ (any number) $times 5$ (any number) $times 4$ (any number except the first two) = $100$?

permutations combinations

share|cite|improve this question

edited Mar 22 at 5:17

MarianD

2,2101618

asked Mar 22 at 4:08

saturnstrollsaturnstroll

11

$endgroup$

closed as off-topic by John Douma, Leucippus, Lord Shark the Unknown, egreg, Eevee Trainer Mar 24 at 4:46

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." – John Douma, Leucippus, Lord Shark the Unknown, egreg, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

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  $begingroup$
  How many choices for first digit? Second digit? Try to work it out, you will get the answer.
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 22 at 4:11
  

  
  
  
  


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  $begingroup$
  I just watched this video: youtube.com/watch?v=Dw231YQi9yE
  $endgroup$
  – saturnstroll
  Mar 22 at 4:15
  

  
  
  
  


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  $begingroup$
  There are $5^3$ such combinations possible.
  $endgroup$
  – Dbchatto67
  Mar 22 at 4:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  that answered the question. but now I'd like to add some limitations. ie: I don't want more than 2 repeats per 3 combo. so, is it correct to say 5 (any number) x 5 (any number) x 4 (any number except the first two) = 100 ?
  $endgroup$
  – saturnstroll
  Mar 22 at 4:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Add it on the body of the question what you have said in the above comment.
  $endgroup$
  – Dbchatto67
  Mar 22 at 4:18
  

  
  
  
  

 | 
show 6 more comments

0

$begingroup$

The question is:

Using the numbers $1, 2, 3, 4,$ and $5$, how many $3$-digit combinations can you make? Repetitions are allowed. Thanks!

Edit:

I just watched the video Total 3 Digit Numbers If Repetition allowed & Not allowed - Permutations & Combinations Problems. That answered my question.

But now I'd like to add some limitations, i. e.: I don't want more than $2$ repeats per $3$ combo. So, is it correct to say $5$ (any number) $times 5$ (any number) $times 4$ (any number except the first two) = $100$?

permutations combinations

share|cite|improve this question

edited Mar 22 at 5:17

MarianD

2,2101618

asked Mar 22 at 4:08

saturnstrollsaturnstroll

11

$endgroup$

closed as off-topic by John Douma, Leucippus, Lord Shark the Unknown, egreg, Eevee Trainer Mar 24 at 4:46

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." – John Douma, Leucippus, Lord Shark the Unknown, egreg, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  How many choices for first digit? Second digit? Try to work it out, you will get the answer.
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 22 at 4:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just watched this video: youtube.com/watch?v=Dw231YQi9yE
  $endgroup$
  – saturnstroll
  Mar 22 at 4:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are $5^3$ such combinations possible.
  $endgroup$
  – Dbchatto67
  Mar 22 at 4:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  that answered the question. but now I'd like to add some limitations. ie: I don't want more than 2 repeats per 3 combo. so, is it correct to say 5 (any number) x 5 (any number) x 4 (any number except the first two) = 100 ?
  $endgroup$
  – saturnstroll
  Mar 22 at 4:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Add it on the body of the question what you have said in the above comment.
  $endgroup$
  – Dbchatto67
  Mar 22 at 4:18
  

  
  
  
  

 | 
show 6 more comments

$begingroup$

The question is:

Using the numbers $1, 2, 3, 4,$ and $5$, how many $3$-digit combinations can you make? Repetitions are allowed. Thanks!

Edit:

I just watched the video Total 3 Digit Numbers If Repetition allowed & Not allowed - Permutations & Combinations Problems. That answered my question.

But now I'd like to add some limitations, i. e.: I don't want more than $2$ repeats per $3$ combo. So, is it correct to say $5$ (any number) $times 5$ (any number) $times 4$ (any number except the first two) = $100$?

permutations combinations

share|cite|improve this question

edited Mar 22 at 5:17

MarianD

2,2101618

asked Mar 22 at 4:08

saturnstrollsaturnstroll

11

$endgroup$

share|cite|improve this question

edited Mar 22 at 5:17

MarianD

2,2101618

asked Mar 22 at 4:08

saturnstrollsaturnstroll

11

share|cite|improve this question

edited Mar 22 at 5:17

MarianD

2,2101618

asked Mar 22 at 4:08

saturnstrollsaturnstroll

11

edited Mar 22 at 5:17

MarianD

2,2101618

edited Mar 22 at 5:17

MarianD

2,2101618

asked Mar 22 at 4:08

saturnstrollsaturnstroll

11

asked Mar 22 at 4:08

saturnstrollsaturnstroll

11

1 Answer
1

active

oldest

votes

1

$begingroup$

Original question

You can choose any of the 5 numbers as your first digit (5 options). For your second digit, you can also choose any of the 5 because repetitions are allowed. This gives $5cdot5$ possibilities. Add in a third digit, once again choosing from the 5 numbers, and you have $5cdot5cdot5$, or 125, possibilities.

Question with limitations

Consider two cases. In one case, the first two digits are different. There are $5cdot4$ possibilities that satisfy this. The third can be any of the 5, so there are $5cdot4cdot5=100$ possibilities for the first case.

In the second case, the first two digits are the same. There are $5cdot1=5$ ways this can occur. In this case, the third digit must be different, giving you only 4 options to chose from. This means the second case has $5cdot1cdot4=20$ possibilities.
Adding together the two cases, there are 120 possibilities.

An alternate way of thinking about the problem with limitations is to work backwards from the problem with no limitations. We have 125 possibilities before introducing limitations. Of those possibilities, there are only 5 that violate the limitations (111, 222, 333, 444, and 555). $125-5=120$, so there are 120 possibilities.

share|cite|improve this answer

edited Mar 22 at 23:02

answered Mar 22 at 4:57

Jacob JonesJacob Jones

14111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, Jacob! I follow you. I think there's a typo in the first case scenario, though: "The third can be any of the 5, so there are 5⋅5⋅5=100 possibilities for the first case." Should this be: "The third can be any of the 5, so there are 5⋅4⋅5=100 possibilities for the first case." ? (changed the 5 to a 4)
  $endgroup$
  – saturnstroll
  Mar 22 at 20:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saturnstroll Yes, thank you :)
  $endgroup$
  – Jacob Jones
  Mar 22 at 23:01
  

  
  
  
  

add a comment | 

1

$begingroup$

Original question

You can choose any of the 5 numbers as your first digit (5 options). For your second digit, you can also choose any of the 5 because repetitions are allowed. This gives $5cdot5$ possibilities. Add in a third digit, once again choosing from the 5 numbers, and you have $5cdot5cdot5$, or 125, possibilities.

Question with limitations

Consider two cases. In one case, the first two digits are different. There are $5cdot4$ possibilities that satisfy this. The third can be any of the 5, so there are $5cdot4cdot5=100$ possibilities for the first case.

In the second case, the first two digits are the same. There are $5cdot1=5$ ways this can occur. In this case, the third digit must be different, giving you only 4 options to chose from. This means the second case has $5cdot1cdot4=20$ possibilities.
Adding together the two cases, there are 120 possibilities.

An alternate way of thinking about the problem with limitations is to work backwards from the problem with no limitations. We have 125 possibilities before introducing limitations. Of those possibilities, there are only 5 that violate the limitations (111, 222, 333, 444, and 555). $125-5=120$, so there are 120 possibilities.

share|cite|improve this answer

edited Mar 22 at 23:02

answered Mar 22 at 4:57

Jacob JonesJacob Jones

14111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, Jacob! I follow you. I think there's a typo in the first case scenario, though: "The third can be any of the 5, so there are 5⋅5⋅5=100 possibilities for the first case." Should this be: "The third can be any of the 5, so there are 5⋅4⋅5=100 possibilities for the first case." ? (changed the 5 to a 4)
  $endgroup$
  – saturnstroll
  Mar 22 at 20:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saturnstroll Yes, thank you :)
  $endgroup$
  – Jacob Jones
  Mar 22 at 23:01
  

  
  
  
  

add a comment | 

1

$begingroup$

Original question

You can choose any of the 5 numbers as your first digit (5 options). For your second digit, you can also choose any of the 5 because repetitions are allowed. This gives $5cdot5$ possibilities. Add in a third digit, once again choosing from the 5 numbers, and you have $5cdot5cdot5$, or 125, possibilities.

Question with limitations

Consider two cases. In one case, the first two digits are different. There are $5cdot4$ possibilities that satisfy this. The third can be any of the 5, so there are $5cdot4cdot5=100$ possibilities for the first case.

In the second case, the first two digits are the same. There are $5cdot1=5$ ways this can occur. In this case, the third digit must be different, giving you only 4 options to chose from. This means the second case has $5cdot1cdot4=20$ possibilities.
Adding together the two cases, there are 120 possibilities.

An alternate way of thinking about the problem with limitations is to work backwards from the problem with no limitations. We have 125 possibilities before introducing limitations. Of those possibilities, there are only 5 that violate the limitations (111, 222, 333, 444, and 555). $125-5=120$, so there are 120 possibilities.

share|cite|improve this answer

edited Mar 22 at 23:02

answered Mar 22 at 4:57

Jacob JonesJacob Jones

14111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, Jacob! I follow you. I think there's a typo in the first case scenario, though: "The third can be any of the 5, so there are 5⋅5⋅5=100 possibilities for the first case." Should this be: "The third can be any of the 5, so there are 5⋅4⋅5=100 possibilities for the first case." ? (changed the 5 to a 4)
  $endgroup$
  – saturnstroll
  Mar 22 at 20:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saturnstroll Yes, thank you :)
  $endgroup$
  – Jacob Jones
  Mar 22 at 23:01
  

  
  
  
  

add a comment | 

$begingroup$

Original question

You can choose any of the 5 numbers as your first digit (5 options). For your second digit, you can also choose any of the 5 because repetitions are allowed. This gives $5cdot5$ possibilities. Add in a third digit, once again choosing from the 5 numbers, and you have $5cdot5cdot5$, or 125, possibilities.

Question with limitations

Consider two cases. In one case, the first two digits are different. There are $5cdot4$ possibilities that satisfy this. The third can be any of the 5, so there are $5cdot4cdot5=100$ possibilities for the first case.

In the second case, the first two digits are the same. There are $5cdot1=5$ ways this can occur. In this case, the third digit must be different, giving you only 4 options to chose from. This means the second case has $5cdot1cdot4=20$ possibilities.
Adding together the two cases, there are 120 possibilities.

An alternate way of thinking about the problem with limitations is to work backwards from the problem with no limitations. We have 125 possibilities before introducing limitations. Of those possibilities, there are only 5 that violate the limitations (111, 222, 333, 444, and 555). $125-5=120$, so there are 120 possibilities.

share|cite|improve this answer

edited Mar 22 at 23:02

answered Mar 22 at 4:57

Jacob JonesJacob Jones

14111

$endgroup$

share|cite|improve this answer

edited Mar 22 at 23:02

answered Mar 22 at 4:57

Jacob JonesJacob Jones

14111

answered Mar 22 at 4:57

Jacob JonesJacob Jones

14111

answered Mar 22 at 4:57

Jacob JonesJacob Jones

14111