Answer is d. The first digit can be 1-9 so 9 and 5 more digit 10^5. The 7 digit number's first digit can be 1-9 and other 6 digit could be 10^6.

9x10^6 - 9x10^5 = 9x 10^5 (10-9) = 9x9 x10^5 =81 x 10^5

The question basically asks:
(Total number of possible combinations if you have 7 slots) - (Total number of possible combinations if you have 6 slots).

Number of available digits: 10 (0-9)
Restriction: First number cannot be 0

When we have 7 slots, i.e., __.__.__.__.__.__.__ (Visualization always helps)
Using Fundamental Counting Principal, we have \(9*10*10*10*10*10*10\) = \(9*10^6\) possibilities

Similarly when we have 6 slots, we have \(9*10^5\) possibilities

Difference:
(\(9*10^6\)) - (\(9*10^5\)) = (\(9*10^5\))(10-1) = (\(9*10^5\))(9) = \(81*10^5\)

Answer: D

Kindly correct me if I'm wrong. Thanks!

6 - digits

9 x 10 x 10 x 10 x 10 x 10 <---- 9 options for digit 1 (since 0 cannot be the first) AND 10 options for each of the other (5) digits

7 - digits

9 x 10 x 10 x 10 x 10 x 10 x 10 <---- 9 options for digit 1 (since 0 cannot be the first) AND 10 options for each of the other (6) digits

(9 x 10^6) - (9 x 10^5)
= (9 x 10^5) (10^1 - 1)
= 9 x 9 x 10^5
= 81 x 10^5

Answer is D.
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The problem can be solved in two ways:
A. If only 6 digits, including a '0', are used
B. All digits are used.

B seems to be the case here, however.

A. 5 digits can take the 1st place and each of the rest 5 places can be taken in 6 ways.
Ways a telephone number can be formed = \(5*6^5\)

In case of a 7 digit number, total ways are \(5*6^6\)
Difference = \(5*6^6 - 5*6^5 = 5^2*6^5\)

B. 9 digits can take the 1st place and each of the rest 5 places can be taken in 10 ways.
Ways a telephone number can be formed = \(9*10^5\)

In case of a 7 digit number, total ways are \(9*10^6\)
Difference = \(9*10^6 - 9*10^5 = 9^2*10^5\)

Answer D.
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