How many different phone numbers are possible with 2 certain numbers? [on hold]

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Assume all telephone numbers are 10 digits long, consisting of a 3-digit area code, then a 3-digit "exchange" number, followed by a 4-digit number. How many telephone numbers have no 0 in the area code and no 9 in the final 4-digit number?



I want to say the answer is $8^10$, but will that account for 0 also being the second or third number in the area code and 9 being the second, third or fourth number?




combinatorics discrete-mathematics




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put on hold as off-topic by BLAZE, Jyrki Lahtonen, amWhy, José Carlos Santos, Key Flex 2 days ago


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – BLAZE, Jyrki Lahtonen, amWhy, José Carlos Santos, Key Flex
If this question can be reworded to fit the rules in the help center, please edit the question.








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    What have you tried? Posting homework questions with no context or attempt is not likely to be met with enthusiasm. In order to receive less downvotes, please consider using MathJax.
    – Prime
    2 days ago















up vote
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down vote

favorite












Assume all telephone numbers are 10 digits long, consisting of a 3-digit area code, then a 3-digit "exchange" number, followed by a 4-digit number. How many telephone numbers have no 0 in the area code and no 9 in the final 4-digit number?



I want to say the answer is $8^10$, but will that account for 0 also being the second or third number in the area code and 9 being the second, third or fourth number?




combinatorics discrete-mathematics




share|cite|improve this question











put on hold as off-topic by BLAZE, Jyrki Lahtonen, amWhy, José Carlos Santos, Key Flex 2 days ago


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – BLAZE, Jyrki Lahtonen, amWhy, José Carlos Santos, Key Flex
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    What have you tried? Posting homework questions with no context or attempt is not likely to be met with enthusiasm. In order to receive less downvotes, please consider using MathJax.
    – Prime
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Assume all telephone numbers are 10 digits long, consisting of a 3-digit area code, then a 3-digit "exchange" number, followed by a 4-digit number. How many telephone numbers have no 0 in the area code and no 9 in the final 4-digit number?



I want to say the answer is $8^10$, but will that account for 0 also being the second or third number in the area code and 9 being the second, third or fourth number?




combinatorics discrete-mathematics




share|cite|improve this question











Assume all telephone numbers are 10 digits long, consisting of a 3-digit area code, then a 3-digit "exchange" number, followed by a 4-digit number. How many telephone numbers have no 0 in the area code and no 9 in the final 4-digit number?



I want to say the answer is $8^10$, but will that account for 0 also being the second or third number in the area code and 9 being the second, third or fourth number?





combinatorics discrete-mathematics





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked 2 days ago







user581600











put on hold as off-topic by BLAZE, Jyrki Lahtonen, amWhy, José Carlos Santos, Key Flex 2 days ago


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – BLAZE, Jyrki Lahtonen, amWhy, José Carlos Santos, Key Flex
If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by BLAZE, Jyrki Lahtonen, amWhy, José Carlos Santos, Key Flex 2 days ago


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – BLAZE, Jyrki Lahtonen, amWhy, José Carlos Santos, Key Flex
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    What have you tried? Posting homework questions with no context or attempt is not likely to be met with enthusiasm. In order to receive less downvotes, please consider using MathJax.
    – Prime
    2 days ago













  • 1




    What have you tried? Posting homework questions with no context or attempt is not likely to be met with enthusiasm. In order to receive less downvotes, please consider using MathJax.
    – Prime
    2 days ago








1




1




What have you tried? Posting homework questions with no context or attempt is not likely to be met with enthusiasm. In order to receive less downvotes, please consider using MathJax.
– Prime
2 days ago





What have you tried? Posting homework questions with no context or attempt is not likely to be met with enthusiasm. In order to receive less downvotes, please consider using MathJax.
– Prime
2 days ago











2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










So let's work with the area code first consisting of 3-digits. If we had no constraint, then the total possible number of area codes is clearly $10^3 = 1000$ (starting from 000 up till 999).



Since no $0$ is allowed in the Area code ($AAA$), then we have nine allowed digits, hence a total number of area codes equal to $9^3$. So,
$$textTotal number of area codes = 9^3$$



Similarly, the last digits has no 9 and using the same reasoning above, the total is
$$textTotal number of last 4-digits codes = 9^4$$



The exchange number is left intact (no constraints) so:
$$textTotal number of last exchange numbers = 10^3$$



So, the total number of possible phone numbers will be $9^3 times 9^4 times 10^3$






share|cite|improve this answer





















  • I get it now thank you for the thorough explanation!
    – user581600
    2 days ago










  • If you found the explanation useful, you could mark the answer as correct so as to assist further readers.
    – Ahmad Bazzi
    2 days ago

















up vote
0
down vote













  • $9^3$ for the area code ( no zero)

  • $10^3$ for the exchange

  • $9^4$ for the last 4 digits (no 9)

Total $= 9^7 cdot 10^3$






share|cite|improve this answer





















  • Ohh now I understand! Thank you!
    – user581600
    2 days ago
















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










So let's work with the area code first consisting of 3-digits. If we had no constraint, then the total possible number of area codes is clearly $10^3 = 1000$ (starting from 000 up till 999).



Since no $0$ is allowed in the Area code ($AAA$), then we have nine allowed digits, hence a total number of area codes equal to $9^3$. So,
$$textTotal number of area codes = 9^3$$



Similarly, the last digits has no 9 and using the same reasoning above, the total is
$$textTotal number of last 4-digits codes = 9^4$$



The exchange number is left intact (no constraints) so:
$$textTotal number of last exchange numbers = 10^3$$



So, the total number of possible phone numbers will be $9^3 times 9^4 times 10^3$






share|cite|improve this answer





















  • I get it now thank you for the thorough explanation!
    – user581600
    2 days ago










  • If you found the explanation useful, you could mark the answer as correct so as to assist further readers.
    – Ahmad Bazzi
    2 days ago














up vote
0
down vote



accepted










So let's work with the area code first consisting of 3-digits. If we had no constraint, then the total possible number of area codes is clearly $10^3 = 1000$ (starting from 000 up till 999).



Since no $0$ is allowed in the Area code ($AAA$), then we have nine allowed digits, hence a total number of area codes equal to $9^3$. So,
$$textTotal number of area codes = 9^3$$



Similarly, the last digits has no 9 and using the same reasoning above, the total is
$$textTotal number of last 4-digits codes = 9^4$$



The exchange number is left intact (no constraints) so:
$$textTotal number of last exchange numbers = 10^3$$



So, the total number of possible phone numbers will be $9^3 times 9^4 times 10^3$






share|cite|improve this answer





















  • I get it now thank you for the thorough explanation!
    – user581600
    2 days ago










  • If you found the explanation useful, you could mark the answer as correct so as to assist further readers.
    – Ahmad Bazzi
    2 days ago












up vote
0
down vote



accepted







up vote
0
down vote



accepted






So let's work with the area code first consisting of 3-digits. If we had no constraint, then the total possible number of area codes is clearly $10^3 = 1000$ (starting from 000 up till 999).



Since no $0$ is allowed in the Area code ($AAA$), then we have nine allowed digits, hence a total number of area codes equal to $9^3$. So,
$$textTotal number of area codes = 9^3$$



Similarly, the last digits has no 9 and using the same reasoning above, the total is
$$textTotal number of last 4-digits codes = 9^4$$



The exchange number is left intact (no constraints) so:
$$textTotal number of last exchange numbers = 10^3$$



So, the total number of possible phone numbers will be $9^3 times 9^4 times 10^3$






share|cite|improve this answer













So let's work with the area code first consisting of 3-digits. If we had no constraint, then the total possible number of area codes is clearly $10^3 = 1000$ (starting from 000 up till 999).



Since no $0$ is allowed in the Area code ($AAA$), then we have nine allowed digits, hence a total number of area codes equal to $9^3$. So,
$$textTotal number of area codes = 9^3$$



Similarly, the last digits has no 9 and using the same reasoning above, the total is
$$textTotal number of last 4-digits codes = 9^4$$



The exchange number is left intact (no constraints) so:
$$textTotal number of last exchange numbers = 10^3$$



So, the total number of possible phone numbers will be $9^3 times 9^4 times 10^3$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered 2 days ago









Ahmad Bazzi

2,162417




2,162417











  • I get it now thank you for the thorough explanation!
    – user581600
    2 days ago










  • If you found the explanation useful, you could mark the answer as correct so as to assist further readers.
    – Ahmad Bazzi
    2 days ago
















  • I get it now thank you for the thorough explanation!
    – user581600
    2 days ago










  • If you found the explanation useful, you could mark the answer as correct so as to assist further readers.
    – Ahmad Bazzi
    2 days ago















I get it now thank you for the thorough explanation!
– user581600
2 days ago




I get it now thank you for the thorough explanation!
– user581600
2 days ago












If you found the explanation useful, you could mark the answer as correct so as to assist further readers.
– Ahmad Bazzi
2 days ago




If you found the explanation useful, you could mark the answer as correct so as to assist further readers.
– Ahmad Bazzi
2 days ago










up vote
0
down vote













  • $9^3$ for the area code ( no zero)

  • $10^3$ for the exchange

  • $9^4$ for the last 4 digits (no 9)

Total $= 9^7 cdot 10^3$






share|cite|improve this answer





















  • Ohh now I understand! Thank you!
    – user581600
    2 days ago














up vote
0
down vote













  • $9^3$ for the area code ( no zero)

  • $10^3$ for the exchange

  • $9^4$ for the last 4 digits (no 9)

Total $= 9^7 cdot 10^3$






share|cite|improve this answer





















  • Ohh now I understand! Thank you!
    – user581600
    2 days ago












up vote
0
down vote










up vote
0
down vote









  • $9^3$ for the area code ( no zero)

  • $10^3$ for the exchange

  • $9^4$ for the last 4 digits (no 9)

Total $= 9^7 cdot 10^3$






share|cite|improve this answer













  • $9^3$ for the area code ( no zero)

  • $10^3$ for the exchange

  • $9^4$ for the last 4 digits (no 9)

Total $= 9^7 cdot 10^3$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered 2 days ago









WW1

6,2121712




6,2121712











  • Ohh now I understand! Thank you!
    – user581600
    2 days ago
















  • Ohh now I understand! Thank you!
    – user581600
    2 days ago















Ohh now I understand! Thank you!
– user581600
2 days ago




Ohh now I understand! Thank you!
– user581600
2 days ago


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