This answer is based on assumptions that indicate we want the size of a set S of numbers, the five smallest of which are 1,111,111
1,111,112
1,111,114
1,111,117
1,111,121
and the two largest of which are 7,777,774
7,777,777
The following two numbers would therefore not be in S: 1
777,777 The answer is: 47 = (4x4x4x4x4x4x4) = 16,384
First digit can be any of nine (not "0"), second can also be any of nine, third any of eight, fourth any of seven and fifth any of six, that is 9 x 9 x 8 x 7 x 6 = 27216 such numbers.
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.
The digit 7 is in the tenths place, therefore its value is 7/10 (or seven tenths).
Best way to answer this question is by figuring out the probability of a phone number not containing the number 7. There are 9 other numbers that are not a 7, so odds of each digit not being a 7 is 90%. For a 7-digit number, calculation would be 0.9*0.9*0.9*0.9*0.9*0.9*0.9 or 0.9^7. This equals 47.8%, which means the odds of a phone number containing at least one 7 is 100% - 47.8%, which is 52.2%. But phone numbers do not begin with 0 or 1, so if we account for that, the odds of the first digit not being a 7 is 7/8, not 9/10. We replace the first 0.9 with 0.875, and the result is 53.5%. There are other scenarios we can take out, like numbers that begin with 555, but I'll just leave it at that.
There are no seven diamonds below.
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.
seven which are 146 164 416 461 614 641
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."
Seven of them.
252
The number of possibilities is 9 x 10 x 10 x 10 x 10 x 10 x 9 = 8,100,000.
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.
Same as the odd numbers anywhere else, except the first digit is always "7"