There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.
You have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.
# 230 # 203 # 320 # 302
If the number can contain repeated digits, the answer is 800000. Without repetition, there are 483840.
36 two digit numbers can be formed...(:From Rafaelrz: The question can be stated as;how many permutations of two different digits can beobtained from a set of six different digits ?Answer:nPr equals n!/(n-r) ...... for n = 6, r = 26P2 equals 6!/(6-2)! equals 30 Permutations.
With repetition, 94 = 6561. Without repetition, 9*8*7*6 = 3024.
There are 7,290 different 4-digit numbers that can be formed from the digits 1-9 without repetition.
30 without repetition (6P2) 66 with repetition (12C2)
15 of them.
I take it that you want to make three digits numbers with 8,7,3, and 6 without repetition. The first digit cane be selected from among 4 digits, the second from 3 digits, the third digit from 2, hence the number of three digit numbers that can be formed without repetition is 4 x 3 x 2 = 24
To calculate the number of four-digit numbers that can be made using the digits 1, 4, 5, and 9 without repetition, we use the permutation formula. Since there are 4 digits to choose from for the first digit, 3 for the second, 2 for the third, and 1 for the fourth, the total number of permutations is 4 x 3 x 2 x 1 = 24. Therefore, there are 24 different four-digit numbers that can be formed using the digits 1, 4, 5, and 9 without repetition.
Six (6)
The four digits can be used to produce infinitely many different numbers if repetition is permitted. Without repetition, there are 24 possible numbers. A lot more can be produced if the numbers are combined using binary oprations, fore example, 19 * 8/4 = 19*2 = 38.
There are 10 to the 10th power possibilities of ISBN numbers if d represents a digit from 0 to 9 and repetition of digits are allowed. That means there are 10,000,000,000 ISBN numbers possible.
You have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.
9*8*7 = 504 of them.
3*2*1 = 6 of them.
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.