Nope, look at 1000, this is a 4 digit number and we can't list it with 1,2,3, and 4
I would not like a list all possible 4 digit combination using 0-9.
There are five such numbers: 11, 12, 15, 24 and 36.
There are ten thousand possible pin numbers with four digits. To generate your list, start at 0000 and start counting, to 9999.
-1
To make this simpler, I'll assume that numbers with a leading zero like 0179 are "four digit numbers" for the purposes of this question. The number of such numbers that can be formed by those numbers without repeating one of the digits is 4x3x2x1 = 24. We could list them all and add them up, but let's instead just realize that we can choose any digit and put it in any position, and the other digits can then be arranged in 3x2x1 = 6 ways. So the sum for any position is (0+1+7+9)*6 = 102, and the sum of the set of "four digit" numbers is 102x1000+102x100+102x10+102 = 113322. You can figure out for yourself what the sum would be if you exclude the numbers that are really three digit numbers (hint: use the technique above to find the sum of the three digit numbers that can be formed using the digits 1, 7, and 9; then subtract that from the other total).
1,000. The list looks just like the counting numbers from 000 to 999 .
Too many to list. Allowing leading zeros, there are 5040 such numbers. 4536 without leading zeros.
There are only six: 247, 274, 427, 472, 724 and 742.
Just do it systematically, START WITH 0123 0124 0125 etc 0129 now 1234 1235 keep going! There are 10 numbers from 0 to 9. So the number of combinations of 4 digits numbers is 30. 10C4 = 10!/[(10 - 4)!4!] = 10!/(6!4!) = (10 x 9 x 8 x 7 x 6!)/(6! x 4 x 3 x 2 x 1)= 10 x 3 = 30
I would not like a list all possible 4 digit combination using 0-9.
They belong to a set of numbers whose first digit is 8. There are 720 such numbers and I regret that I do not have the patience to list them all.
This is a problem involving combinations. We know we need a 4 digit number. So I'll write 4 blank spaces, to symbolize each digit. ___ ___ ___ ___ The number also has to be even. That means that the last digit must be even. (2, 4, 6, 8, 0) Your list of numbers is 1, 2, 3, 5, 6, 8, 0. How many numbers from this list will work in the last digit place? 2, 6, 8, 0. -- 4 numbers. ___ ___ ___ _4_ For the third digit, any number will work. We have 7 choices. ___ ___ _7_ _4_ for the 2nd digit, the same applies. any number from your list will work. ___ _7_ _7_ _4_ For the first digit, only 6 will work. if you put a 0 in the first digit place, it becomes a 3 digit number. _6_ _7_ _7_ _4_ Now, all we do it multiply these numbers together. 6 * 7 * 7 * 4 = 1176. This means, we can create 1176 unique 4 digit numbers that are even with the list of numbers available. 1000 1002 1006 1008 1010 1012 etc.
99. Each of the numbers in the list can only be used by itself otherwise you would get a 4-digit (or longer) number.
1,000 of them. The list of possibilities will look exactly like the counting numbers from 000 to 999 .
There are 12,500 of them and I am not going to list them. One of them is 200000.
There are five such numbers: 11, 12, 15, 24 and 36.
There is no such thing as "intergers".The complete list of two-digit positive whole numbers lists 90 of them.