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A common problem that says to make $8$-digit phone numbers, assuming that the first digit can’t be a $0$ or a $3$? Re-solve the previous problem, except now assume also that the phone number is not allowed to start with $433$.

$\\$I have solved both of the problems and especially the second one after a wrong trial. I am trying to find out why the first trial was wrong.

Here is what I did to solve the second problems in first trial.

If the number is not starting with $4$ then there are $9$ ways to choose digits other than $4$ and same argument for not selecting $3$'s and for the rest part there are $10$ ways to select any digit from $0-9$.The total number of way, becomes $9 \times 9 \times 9 \times 10^5=\ 72900000$, which is wrong. I would like to understand why this approach is not correct or why a direct multiplication cannot be applied to solve this problem. I hate to ask such elementary question but I want to understand the reason to fail to solve the problem at first attempt. Any explanation is highly appreciated.

vbm
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    You arguably could use a direct multiplication approach., but it would be that rather than picking the first second and third digits separately we pick the first three digits simultaneously. There are $1000-1$ valid ways to pick the first three digits simultaneously giving $999\times 10^5$. Do not confuse the statement "Not allowed to start with 433" with the statement "The first digit is not allowed to be a $4$ regardless any other chosen digits while the second digit is not allowed to be a $3$ regardless any other chosen digits, etc..." 43712345 is a valid phone number. – JMoravitz Oct 08 '19 at 16:37
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    Your argument counts the number of ways to avoid $4$ as the first digit and $3$ as the second digit and $3$ as the third digit. But it won't count acceptable phone numbers, such as those that begin with $333$. – Robert Shore Oct 08 '19 at 16:38

3 Answers3

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The correct answer should exclude only those numbers where the first digit was $4$ AND the second digit was $3$ AND the third digit was $3$.

Your answer excludes all numbers where the first digit was $4$ OR the second digit was $3$ OR the third digit was $3$.

  • I see almost everyone suggesting to use $OR$ but would you please clarify the reason for not using $AND$ in the calculation? This is the area where I get confused to choose an 'OR' or an 'and'. – vbm Oct 08 '19 at 16:44
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    For example, you excluded the numbers such as $41123456$ (because of leading $4$), numbers such as $73956721$ (because of second $3$) and $55399011$ (because of the third $3$) - none of them beginning with $433$. –  Oct 08 '19 at 16:57
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In your solution you're assuming that the first digit isn't 4, the second digit isn't 3, and the third digit isn't 3. That and should be an or.

In particular, your answer is not counting the number $43000000$.

79037662
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The problem with your approach is that you are including the other invalid phone number possibilities which start with $0$ and $3$. The best approach would be to find the number of valid phone numbers not starting with $0$ and $3$ and then subtract possibilities with $433$.

Possibilities where number does not start with $0$ or $3$ $=8\times 10^7$

Possibilities where number does not start with $433$ $=10^5$

Total Possibilities $=80,000,000-100,000=79,900,000$

Sam
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