How many numbers from 0 to 9999 contain the digit 7 exactly once?

Therefore, 280 numbers contain the digit 7 between 0 and 999.

How many integer numbers between 0 and 9999 have exactly one digit 1 and exactly one digit 3?

Hence the number of integers between 0 and 9999 that have exactly one digit 1 and exactly one digit 3 is 4⋅3⋅8⋅8=768.

What is the probability that an integer between 0 and 9999 has exactly one 8and one 9?

The probability is 768/10000.

How many integers strictly from 0 and 10000 have exactly one digit equal to 5?

There are 4 places to put the 5. There are 9 choices (0-4 and 6-9) for remaining 3 digits, so they can be chosen in 9^3 = 729 ways. The total is 4 * 729 = 2916. pls mark as brainliest answer !!!

How many number of times will the digit 9 be written when listing the integers from 1 to 10000?

Originally Answered: When we write from 1 to 10000, how many times is the digit 9 written? 4000 times. From 1 to 10000 = 300*10 + 1000 (9000,9001,9002….) = 4000 times.

How many numbers between 0 and 10000 have a sum of digits equal to 5?

So, we can state that 2916 numbers between 0 and 10000 have one and only one digit as 5.

How many positive integers less than 1000000 have ex actly one digit equal to 9 and have a sum of digits equal 13?

How many positive integers <1,000,000,000 have exactly one digit equal to 9 and have the sum of digits equal to 13? Then 330∗9=2970. So there are 2970 integers that satisfy the problem.

How many integers lies between and 5?

The integers -4, -3, -2, -1, 0, 1, 2, 3 and 4 lie between –5 and 5. So, nine integers lie between –5 and 5.

How many numbers between 1 and 1000 contain the digit 0?

The digit zero appears only in the numbers 10, 20, 30, 40, 50, 60, 70, 80, 90, 100. Thus, there are 11 zeros among the digits. This is the same as the number of digits in writing the numbers from 1 to 10. Now let us look at the numbers from 1 to 1000.

How many possible 4 digit numbers are there with exactly 8 and 9?

So we have: 12*8*8-48 = 720 possible 4 digit numbers. 48 is the result of 46 + 2 from case #2 because we have to subtract outcomes where the first two digits are 0 or the first digit is a 0 eg: 0098. Therefore, the total is 2 + 46 + 720 = 768 numbers with exactly 8 and 9 in it. The probability is 768/10000.

How many 4-digit numbers are there from 1000 to 9999?

You can count from 1000 to 9999 for the answer, but easier to use a formula: The answer is the same for the question: How many 4-digit numbers are there? 1000 is the smallest 4-digit number, and 9999 is the largest.

How many numbers between 0 and 10000 have one and only one?

So, we can state that 2916 numbers between 0 and 10000 have one and only one digit as 5. The numbers looked for must have one and only one digit as 5. Thus, the three other digits may be any of 0, 1, 2, 3, 4, 6, 7, 8, 9. Having each digit nine possible values, a set of three of them can have 9 ∗ 9 ∗ 9 = 9 3 = 729 different values.

How many even and odd numbers are there between 1000 and 9999?

There are 9,000 unique numbers between 1000 and 9999 (which are all four digit numbers that do not start with 0, counting 1000 as part of the set). Half of those number will be even, and the other half odd.