Q:

How many 7-digit phone numbers are possible, assuming that the first digit can’t be a 0 or a 1

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. 

A:

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
$$|B\times A^6| = |B||A^6|=|B||A|^6 = 8\cdot10^6.$$

A:

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)

