I use the multiplication rule. Because that the very first digit I have 8 choices. For the last 6 number I have actually 10 options for each. Therefore answer is $8 \cdot 10 ^6$.

You are watching: How many number combinations in 7 digits

Is there any kind of other means to solve this problems. I usually acquire a lot of understanding from solving the problems in different ways. Please compose which theorems etc. You have used.


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Another way to look in ~ 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 require to add $1$ as we counting $2000000$ as a valid number)


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What you have said is basically correct, but perhaps a an ext "formal" means (which is where the multiplication ascendancy comes from) is to construct the set of all possibilities, and also 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 feasible phone numbers is $B \times A^6$, and therefore the variety of possible phone number is$$|B\times A^6| = |B||A^6|=|B||A|^6 = 8\cdot10^6.$$


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

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

number that 7 digit numbers no lead through 0 or 1 : 8,000,000

This more just taking a complement of a set. ( for this reason an interior type of inclusion-exclusion)

You can realize all 7 have at the very least 8, obtain $8^7$ climate realize each of 6 have actually 2 much more with the seventh having $8=2^3$, because that $2^9$ and have fun adding up every $64=2^6$ combinations all together. An ext a property of a powerset i beg your pardon relates come combinations, as the total variety of combinations of every sizes, is the number of distinguishable says which in includes all subsets, (Also tedious)


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