120 times.
1, 5, 8, and 0.
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.
You look at the digit in the one millions position, which is a 4. That causes the number to be rounded down to 530000000.
Mean: Add all of the numbers in the data set, then divide by the amount of numbers in the set of data. Median: Order the numbers from least to greatest and find the middle number. If there is more than one number in the middle, add the 2 numbers together, then divide by two. Mode: To find the mode, look for the number that appears most in the data set. If there is a tie, write them both down. Range: To determine the range, subtract the smallest number to the biggest number.
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.
The digit 3 will be written 20 times.
18 times
10
20, don't forget 90, 91, 92, 93 ...... etc
You line them up and down and then multiply the first ones up and down on the right. then you do the diagonally ones and then you put a zero on the bottom right. then you go to the second number on the right and do the one diagonally and then the on up and down. then you add the two together and you have your answer. example:
Write the two sums down one below the other and then draw a line underneath like so, then just do it backwards. do 4x6 then write it below. If you get a double digit then write the 4 underneath the 6 and a small 2 underneath the 4 next to the 6. Then when you times the next line add it on to the total 2344 x 1946 --------
Then you write down both numbers
118 The following from rjsiekman; Think of it as the number of times the 3 will be written in the hundreds, tens, and ones column. This analysis assumes whole numbers; otherwise you can write a 3 an infinite number of times. In the hundreds, 301.....399 is 99 (because it is stated the numbers between 300 & 400; 300 is not included). In the tens, 3 will be written for 330....339 which adds another 10. In the ones, 3 will be written for 303, 313....393 which adds another 10. So, in total, the 3 will be written 119 times.
Ask your instructor if you can write down your 10 digit code.
11, 12, 13, 14, 15, 16, 17, 18, 19As shown above, the digit 2 is only written once.
In numbers it is: 4,170,000 dollars
Press down the 2 and 3 buttons until you get a 3 digit number. Then write it down. Then press the AM FM button and you will get another 3 digit number. Write that one down and then post both of them (all 6 digits) here.