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Q: What 4 digit numbers when combined look the same upside down?
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Continue Learning about Statistics

If you write down all the numbers from 300 to 400 how many times will you write the digit 3?

120 times.


How many 4 digit combinations can be made with numbers 3 7 9 4?

The number of combinations possible for taking a specified sub-set of numbers, r, from a set, n, isC(n,r) = n!/[r!(n-r)!]In this case, n = 4 and r = 1, soC(n,r) = 4!/1!3!C(n,r) = 24/6C(n,r) = 4Therefore, there are four possible combinations of the numbers 3, 7, 9 and 4.==================================This contributor strongly disagrees, but is leaving the original answer hereso that others can shop and compare.With their commas, parentheses, and factorials, the formulas are certainly impressive.Only the conclusion is wrong.The question specified "4-digit" combinations, so 'n' and 'r' are both 4.Now, to come down off the pedestal and make it understandable as well asformally rigorous ..."Combination" really means different groups of 4 digits that you can select outof the four you gave us. There's only one of those groups.If you actually want to know how many different 4-digit numbers you can makefrom them, then what you want is called "permutations" of the four digits, andyou can think of it this way, without using 'n', 'r', or parentheses or factorials:The first digit can be any one of 4. For each of those . . .The second digit can be any one of the remaining 3. For each of those . . .The third digit can be either one of the remaining 2.So the total number of different ways to line them up is (4 x 3 x 2) = 24 different 4-digit numbers.


How do you round 534307164 to the nearest 10 million?

You look at the digit in the one millions position, which is a 4. That causes the number to be rounded down to 530000000.


How many three digit number combinations are there?

Let us assume you can repeat numbers. That is to say, we allow 101 and are not bothered by the repeat 1, or 111 for that matter. Let's do it first so that the order matters, that is to say 123 is not the same as 321. The possible numbers for the first digit are 0,1,2,...9 so there are 10 of them, Same for the second. Now for the digit "in front", for example in 123 the digit 1, we must decide if we can allow a 0. To make things easy let's say yes for now. Then there are 10 choices for each digits and 10^3 total choices=1000 Now, lets narrow things down just in case we have some restrictions. If exclude 0 from the digit in front, then there are only 9 choices for that place and we have 900 possibilities. Now you said combinations and that usually means that order does not matter. A permutation is where order matters. So if order does not matter 123 is the same as 321 or 213. This means we over counted and now must divided. Each group of 3 numbers such as 123 can be arranged in 6 different ways, which is 3! for example 123 132 213 232 312 321 We we divided by 3! Assuming the leading 0 was allowed we have 1000/3! combinations, if not we have 900/3! combinations.


What is 2.0045 rounded to the nearest thousandth?

2.005