You are using "combination" which means that there are really only 5 ways to combine those numbers since in combination, order is not important. Your question seems to indicate that you want to know the number of "permutations". That is easily calculated by using the factoral of the number of the set. That is written as 5! and is calculated by multiplying 5x4x3x2x1=120. There are 120 different ways to arrange 5 digits.
There are 90000 5-digit numbers.
120
120. There are 6 digits. If we pick the digits in order, there are 6 possible digits for the first digit 5 remaining digits for the second digit 4 remaining digits for the third digit. 6*5*4 = 120.
There are 5460 five digit numbers with a digit sum of 22.
about 1,0000000000000
500
27 three digit numbers from the digits 3, 5, 7 including repetitions.
To find the even two-digit numbers where the sum of the digits is 5, we need to consider the possible combinations of digits. The digits that sum up to 5 are (1,4) and (2,3). For the numbers to be even, the units digit must be 4, so the possible numbers are 14 and 34. Therefore, there are 2 even two-digit numbers where the sum of the digits is 5.
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
5. Count the number of digits from the first non-zero digit to the last non-zero digit.
52488
Three of them.