The answer will depend on how many digits you have at your disposal.
Working with the 10 digits, and barring leading zeros, there are
8,877,690 numbers.
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.
6*5*4*3*2*1=720 possible numbers
-123456785
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.
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.
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.
6*5*4*3*2*1=720 possible numbers
That would be the numbers in the form "32x" (where "x" can be any digit). In other words, ten numbers.
5*4*3 = 60
6 possible 3 digit combonations
The number of six digit numbers that you can make from ten different digits ifrepetitions of same digit on the six digit number is allowed is 1 000 000 numbers(including number 000 000).If no repetitions of the the same digit are allowed then you have:10P6 = 10!/(10-6)! = 151 200 different six digit numbers(six digit permutations form 10 different digits).
To form a 3-digit number using the digits 5, 6, and 8, we can use each digit in any position. Since we have three digits and they can be repeated, we have 3 choices for the first digit, 3 choices for the second digit, and 3 choices for the third digit. Therefore, the total number of 3-digit numbers is (3 \times 3 \times 3 = 27).
By using the 3 digits of a number we can form 3 different two digit numbers. 3C2 = 3!/[(3 - 2)!2!] = 3!/(1!2!) = (3 x 2!)/2! = 3
To form a 5-digit number using the digits from 1 to 9 (without the digit 0) and allowing repetitions, each position in the number can be filled by any of the 9 digits. Therefore, for each of the 5 positions, there are 9 possible choices. Thus, the total number of 5-digit numbers that can be formed is (9^5 = 59,049).
To form a 4-digit number using the digits 2, 6, 9, and 5 without repeating any digit, you can use all four digits. The number of ways to arrange 4 digits is calculated as 4! (4 factorial), which equals 24. Therefore, you can form 24 different 4-digit numbers with the numerals 2695 without repeating any digit.
-123456785
To form a five-digit number using the digits 1, 2, 3, 4, and 5, each digit must be used exactly once. Since there are 5 unique digits, the total number of different five-digit numbers that can be formed is given by the factorial of the number of digits, which is 5! = 120. Therefore, 120 different five-digit numbers can be formed using the digits 12345.