Study guides

☆☆

Q: How many 3 digit numbers are there that contain at least 1 seven?

Write your answer...

Submit

Still have questions?

Related questions

1 million 1,000,000

If the number can contain repeated digits, the answer is 800000. Without repetition, there are 483840.

4,748,472 Confirmed using the following C# function string sTemp; int total = 0; for (int i = 1000000; i < 10000000; i++) { sTemp = i.ToString(); if (sTemp.Contains("7")) { total++; } } label1.Text = total.ToString();

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

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.

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"

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.

People also asked