252
1 million 1,000,000
Four-digit phone numbers were primarily used in the United States before the widespread adoption of area codes and seven-digit dialing. This system was prevalent from the early 20th century until the late 1950s, when the North American Numbering Plan was introduced, requiring the use of three-digit area codes and seven-digit local numbers. The transition to seven-digit dialing became standard as more telephone lines were added and the need for more numbers grew.
Seven of them.
16, 25, 34, 43, 52, 61, 70 Seven of them.
you say one number first then the other one for example 789 you say seven eight and then nine
To find the total number of seven-digit numbers that contain the number seven at least once, we can use the principle of complementary counting. There are a total of 9,999,999 seven-digit numbers in total. To find the number of seven-digit numbers that do not contain the number seven, we can count the number of choices for each digit (excluding seven), which is 9 choices for each digit. Therefore, there are 9^7 seven-digit numbers that do not contain the number seven. Subtracting this from the total number of seven-digit numbers gives us the number of seven-digit numbers that contain the number seven at least once.
If the number can contain repeated digits, the answer is 800000. Without repetition, there are 483840.
1 million 1,000,000
You have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.
420 and 840
In the 1940s, phone numbers typically consisted of a three-digit area code followed by a seven-digit local number, such as "ABC-1234." This format differed from modern phone numbers, which generally have a three-digit area code followed by a seven-digit local number, such as "(123) 456-7890."
Assuming a seven digit whole number: 1,111,111. But if you have to use at least one (2) digit, then 1,111,112.
Seven of them.
Assuming a seven digit whole number: 1,111,111. But if you have to use at least one (2) digit, then 1,111,112.
For me I think it is conceptually easier to think about the probability that the number will contain the digit seven (and the probability that it does not contain the digit 7 is simply one minus the probability that it does). P(number will contain 7) = P(number is in the seven hundreds) + P(number is not in seven hundreds)*[P(number is in the X hundred seventies)+P(number is not in the X hundred seventies)*P(number ends in seven)] So essentially I am considering all of the numbers in the range that start with seven (i.e., are in the seven hundreds), then all of the numbers in the range that aren't in the seven hundreds but have a 7 in the tens place (i.e., the 170s, 270s, etc., and finally all the numbers that don't have a 7 in the hundred or tens place, but that end in 7). Plugging the numbers into my formula above, I get (100/900)+(800/900)*((10/100)+(90/100)(1/10)) = 7/25 is probability that the number does contain a 7, and 1-(7/25)=18/25 is probability that it does not.
That makes:* 8 options for the first digit * 8 options for second digit * 10 options for the third digit * ... etc. Just multiply all the numbers together.
9,000,0009,999,999 is the last 7 digit number1,000,000 is the first 7 digit numberwhen you subtract the first from the last and add one you get the above answer.