Question

A telephone number consists of seven digits, the first three representing the exchange. how many different telephone numbers are possible within the 537 exchange? the answer is .

Answer 1

Since the first three digits are 537 (the exchange), we have 7-3=4 digits left. In each of those digits, there are 10 possible numbers, 0 to 9. As there are 10 numbers possible for the first digit, there are 10 numbers for each number for the second digit, and so on. For example, if the first digit was 1, we would have 1___ (3 blanks) with each blank having 10 possibilities. As there are 10 possibilities for the second number if the number 1 is there, there are 10 possibilities for the second digit for each number possible for the digit, getting 10*10=100 possible combinations with only two numbers. The same applies to the third and fourth blank, as if the second number is 2, there are 100 possibilities for 12__ (2 blanks) to be a possibility, and there are 10 blanks for each possible combination of the first two numbers, getting 10*10*10=10^3=1,000 possible combinations. Repeat the process for the fourth digit, and you end up with 10,000 possible combinations.

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Answer 2

Using the fundamental counting principle, it is found that 10,000 different telephone numbers are possible within the 537 exchange.

The fundamental counting principle states that if for n independent events, there are $n_1, n_2, ..., n_n$ ways to do them, the total number of ways to do them is given by:

$T = n_1n_2...n_n$

In this problem:

• The first 3 digits are fixed, as 537.
• For the last 4 digits, each of them has 10 possible outcomes, and are independent. Thus:

$T = 10 \times 10 \times 10 \times 10 = 10^4 = 10000$

10,000 different telephone numbers are possible within the 537 exchange.

A similar problem is given at https://brainly.com/question/24067651