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.
$\endgroup$3 Answers
$\begingroup$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)
$\endgroup$ 4 $\begingroup$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$$|B\times A^6| = |B||A^6|=|B||A|^6 = 8\cdot10^6.$$
$\endgroup$ 0 $\begingroup$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)
$\endgroup$ 1More in general
‘Cutter’s Way’ (March 20, 1981)
