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I use the multiplication rule. For the first digit I have 8 choices. For the last 6 digits I have 10 choices for each. So answer is $8 \cdot 10 ^6$.

Is there any other way to solve this problems. I usually gain a lot of insight from solving the problems in different ways. Please write which theorems etc. you have used.

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Another way to look at it is to take the highest possible $7$-digit number and subtract the lowest possible: $9,999,999-2,000,000+1=8,000,000$ (we need to add $1$ as we count $2000000$ as a valid number)

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  • $\begingroup$ How does this work. I do not understand. But it works :-) $\endgroup$ – Xenusi Feb 4 at 22:02
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    $\begingroup$ True, but this only happens to work because all of the valid numbers are adjacent. $\endgroup$ – Don Thousand Feb 4 at 22:02
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    $\begingroup$ @Xenusi: this is just counting all whole numbers between two given numbers. $\endgroup$ – Vasya Feb 4 at 22:05
  • $\begingroup$ Somewhat related: Why numbering should start at zero, Dijkstra 1982 $\endgroup$ – ilkkachu Feb 5 at 9:39
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What you have said is basically correct, but perhaps a more "formal" way (which is where the multiplication rule comes from) is to construct the set of all possibilities, and find its cardinality.

Let $A = \{0,1,2,3,4,5,6,7,8,9\}$, and $B = A \smallsetminus \{0,1\}$. Then the set of all possible phone numbers is $B \times A^6$, and therefore the number of possible phone numbers is $$|B\times A^6| = |B||A^6|=|B||A|^6 = 8\cdot10^6.$$

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number of 7 digit numbers ( including leading 0): 10,000,000

number of 7 digit numbers including lead 0 or 1 : - 2,000,000

number of 7 digit numbers not lead by 0 or 1 : 8,000,000

This more just taking a complement of a set. ( so an interior form of inclusion-exclusion)

You could realize all 7 have at least 8, get $8^7$ then realize each of 6 have 2 more with the seventh having $8=2^3$, for $2^9$ and have fun adding up all $64=2^6$ combinations all together. More a property of a powerset which relates to combinations, as the total number of combinations of all sizes, is the number of distinguishable states which in includes all subsets, (Also tedious)

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  • $\begingroup$ That's immediately what I thought when reading the title. Simple answer to a simple question. This should be the accepted answer as anything more complicated is will be of no benefit to this particular question. $\endgroup$ – Robert Tausig Feb 5 at 10:08

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