Question

A telephone exchange in a city uses 7-digit numbers. None of the telephone exchanges starts with digit 0. How many possible numbers the exchange provides to the public?

Answer 1

Hint: Let us consider 7 places for 7-digit number and first place cannot have digit 0 in it.
So, the possible number of digits in first place is 9 i.e. all digits except 0.
Possible number of digits on all remaining places is 10, i.e. possible number of digits on second, third, fourth, fifth, sixth and seventh places is 10.
Thus, multiply all the possibilities to get the answer.

Complete step-by-step answer:
It is given that a telephone exchange in a city uses 7-digit numbers and none of the telephone exchanges starts with digit 0.
Let us consider 7 places for 7-digit numbers and first place cannot have digit 0 in it.
So, the possible number of digits in first place is 9 i.e. all digits except 0.
Now, for second place, the possible number of digits are 10, because all digits can be possible at second place.
So, the possible number of digits in first place is 10 i.e. all digits.
Similarly, the possible number of digits on all remaining places is 10, i.e. possible number of digits on third, fourth, fifth, sixth and seventh places is 10.
Now, these possibilities are independent from each other. So, we will multiply all the above digits to get the answer.
 $\therefore $ Possible numbers of exchange $ = 9 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10$
   $ = 9 \times {10^6}$
Thus, possible numbers the exchange provides to the public is $9 \times {10^6}$ .

Note: If the possible combinations or permutations are independent from each other, we have to multiply all the possibilities. If the possible combinations or permutations depend on each other, we have to add all the possibilities.