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Question: How many 8 digit mobile numbers can be formed if any digit can be repeated and 0 can also start the mobile number?

The answer is $10^{8}$. However, why couldn't $10\times10\times1\times1\times1\times1\times1\times1$ be correct. Numbers $1$-$10$ can still be repeated.

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  • $\begingroup$ can you elaborate your reasoning of your proposed answer? $\endgroup$
    – Siong Thye Goh
    Commented May 24, 2022 at 4:23
  • $\begingroup$ @SiongThyeGoh You can have 1,1,1,1,1,1,1,1; 2,2,1,1,1,1,1,1; 3,3,1,1,1,1,1,1;etc. $\endgroup$
    – 80808learn
    Commented May 24, 2022 at 4:45

1 Answer 1

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Because the number $10\times 10 \times 1 \times 1 \times 1 \times 1\times 1 \times 1$ represents how many mobile numbers there are with the condition that 2 of the figures can take any of the digits and the rest can only take one possible digit.

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  • $\begingroup$ Could you please explain how they arrived at 10^8 I don't get their reasoning. $\endgroup$
    – 80808learn
    Commented May 24, 2022 at 4:46
  • $\begingroup$ Think of the case of having 2 digits. Once a digit is set in the first figure, the second figure can take on any of the 10 possible digits. Since for every digit on the first figure there are 10 possible digits for the second figure, then there are $10\times 10=10^2$ possible 2 digit "mobile" numbers. Similarly, there are $10^8$ possible 8 digit mobile numbers. $\endgroup$
    – Emilio Junoy
    Commented May 24, 2022 at 4:58

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