Unable to Solve Probability Problem on Telephone Numbers. Please Help!

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Mary.V

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This is the only sum I am unable to solve in a problem set:

How many different 7-digit telephone numbers can be written if the digit 0 cannot appear among the first three digits? Remember that digits can be repeated.
 
Hello. Did you try anything? If you completed the other sums, you probably have at least one idea. Where did you get stuck?

Please take a few minutes to familiarize yourself with the forum's guidelines (AKA: Read Before Posting announcement); you may start with this summary. Thank you! :cool:


PS: Your duplicated thread has been deleted.
 
mmm4444bot said:
Hello. Did you try anything? If you completed the other sums, you probably have at least one idea. Where did you get stuck?

Please take a few minutes to familiarize yourself with the forum's guidelines (AKA: Read Before Posting announcement); you may start with this summary. Thank you! :cool:


PS: Your duplicated thread has been deleted.
Click to expand...

I am in the tenth grade, studying Algebra II.

I think multiplying would work : 7 x 6 x 5 x 4 x 3 x 2 x 1, I’m not sure.

This is problem is part of a multi problem set, the other sums in the set are not related to each other.

Please help, in the form of a sample, way in which to solve this problem.

Thanks so much!
 
Mary.V said:
This is the only sum I am unable to solve in a problem set:

How many different 7-digit telephone numbers can be written if the digit 0 cannot appear among the first three digits? Remember that digits can be repeated.
Click to expand...

How many three-digit numbers are there that contain no zeros? How many four-digit number are there, if zero is allowed in any place?
 
In ascending order:
111-0000
111-0001
111-0002
.
.
.
999-9997
999-9998
999-9999
 
Mary.V said:
I am in the tenth grade, studying Algebra II.

I think multiplying would work : 7 x 6 x 5 x 4 x 3 x 2 x 1, I’m not sure.

This is problem is part of a multi problem set, the other sums in the set are not related to each other.

Please help, in the form of a sample, way in which to solve this problem.

Thanks so much!
Click to expand...

You're suggesting that this is a permutation problem (7!); it isn't, because that involves using each item (digit) only once. There is no such requirement here.

But multiplication will be the key. How many ways are there to choose the first digit? Then the second digit? And so on.

Example: how many 3-digit numbers are there, using only the digits 2 through 8? Each digit can be 2, 3, 4, 5, 6, 7, or 8 (7 possibilities); that gives 7 choices for the first digit, times 7 for the second, times 7 for the third, making a total of 7*7*7 = 343 such numbers.
 
Mary.V said:
Will this be the write answer:

7x7x7x7x7x7x7=823,543.

Thanks a a lot for the help.
Click to expand...

It would be right if there were 7 choices for each digit. But there are more than that (and not the same number for every digit). Give it another try.
 
Mary.V said:
Will this be the write answer:

7x7x7x7x7x7x7=823,543.

Thanks a a lot for the help.
Click to expand...
No, that is not the correct answer, but it is an answer correctly recognizing that arithmetic and thinking are more efficient than listing and counting when dealing with large numbers.

It is frequently a huge help to think about a much simpler problem when you are stuck.

How many ways can you choose a string of two decimal digits when you permit repititions.

The answer is not 2 times 2 = 4. We have 00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11 ..., which is more than 4 and far from complete.

You can get the answer to this very simple problem by doing a complete listing and counting, but is there a quicker and easier way?
 
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Mary.V said:
This is the only sum I am unable to solve in a problem set:
How many different 7-digit telephone numbers can be written if the digit 0 cannot appear among the first three digits? Remember that digits can be repeated.
Click to expand...
There are \(\displaystyle 9^3\) ways to choose the first three digits.
There are \(\displaystyle 10^4\) ways to choose the last four digits.
What is the final answer and why?
 
You're correct in using multiplication to "count" the number of phone numbers.

So your method is __ x __ x __ x __ x __ x __ x __ You have to fill in the blanks. (There are 7 positions because it is a 7-digit phone number.)

How many choices do you have to go in the first position? (Remember: You can choose any digit except 0). Put that answer in the first place.

How many choices do you have for the second position? and the third position? (Keeping in mind the restriction.)

Now what about the fourth place? and fifth and sixth and seventh.

Now multiply them all together to get your final answer.
 
I think I got it:

I need to muntiply 9x9x9x10x10x10x10= 7,290,000

Since in the first three numbers zero is not allowed so only (1,2,3,4,5,6,7,8,9) will be part of the first three sets. Thats nine numbers which will be 9x9x9

Next, since zero is allowed in the next four sets so I have these digits to work with (1,2,3,4,5,6,7,8,9,0,) that totals to 10, which makes its 10x10x10x10 for the next four sets.

Wonder what had happed to the grey matter in my head before, thanks a lot for the help !
 
Mary.V said:
I think I got it:

I need to muntiply 9x9x9x10x10x10x10= 7,290,000

Since in the first three numbers zero is not allowed so only (1,2,3,4,5,6,7,8,9) will be part of the first three sets. Thats nine numbers which will be 9x9x9

Next, since zero is allowed in the next four sets so I have these digits to work with (1,2,3,4,5,6,7,8,9,0,) that totals to 10, which makes its 10x10x10x10 for the next four sets.

Wonder what had happed to the grey matter in my head before, thanks a lot for the help !
Click to expand...

Yes, I think you've got it!
 
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