9999
625. Here's how:There are five odd digits {1,3,5,7,9} so our numbers will be made using only these digits. This is the same number of possibilities as if you had a base 5 number system where the allowed digits are {0,1,2,3,4}. For a 1 digit base-5, there are 5 possibilities, for a two digit base-5 there are 5² = 25 possibilities, so for a 4-digit, there are 5^4 = 625possible. This is the same number you will have with only using odd digits.
Two letters = 26*26 = 676 possibilities 4 digits = 104 possibilities (if numbers such as 0000 are permitted). All in all, 6.76*106 possibilities. That should keep Oxnard University going for a few years.
There are 25 possibilities fitting those rules.
-- If you're allowed to repeat digits in one combination,then there are 104 = 10,000 possibilities.-- If all four of the digits must be different, then there are(10 x 9 x 8 x 7) = 5,040 possibilities.
I assume you mean the sum of the digits.If the first digit is 1, the other two digits have to sum 10. There are 9 different possibilities for this, from 1 + 9 to 9 + 1.If the first digit is 2, the other two digits have to sum 9. There are 10 different possibilities for this, 0 + 9 to 9 + 0.Adding it all up, it seems there are 19 such numbers.
The first digit of your password can be any of the six numbers. As repetition is allowed so can the second, third and fourth so total possibilities is 6 x 6 x 6 x 6 ie 1296
ABC A = 1,2,3,4,5,6,7,8,9; 9 possibilities B = 0,1,2,3,4,5,6,7,8,9; 10 possibilities C = 0,1,2,3,4,5,6,7,8,9; 10 possibilities So there are 9*10*10 = 900 numbers with 3 digits.
If the digits may not be repeated, there are 6x5x4x3x2x1 = 720 numbers, If digits can be repeated, the answer is 6x6x6x6x6x6 = 46,656 numbers If zero is ong the available digits but is excluded from the first place. 5x6x6x6x6x6=38,880 numbers
If the two must be different digits, then there 20 possibilities. If they can be the same one, then there are 25 possibilities.
-- If you accept 4-digit numbers with 'zero' as the first digit, thenthere are 24 possibilities, and their sum is 119,988 .-- If the first digit can't be a 'zero', then there are 18 possibilities,and their sum is 115,992 .Note: I didn't repeat any digits.-- If you allow repeated digits, and accept leading zeros, then there are 256 possibilities,and their sum is 1,279,872.-- If you allow repeated digits but not leading zeros, then there are only 192 possibilities,and their sum is 1,247,904.
625. Here's how:There are five odd digits {1,3,5,7,9} so our numbers will be made using only these digits. This is the same number of possibilities as if you had a base 5 number system where the allowed digits are {0,1,2,3,4}. For a 1 digit base-5, there are 5 possibilities, for a two digit base-5 there are 5² = 25 possibilities, so for a 4-digit, there are 5^4 = 625possible. This is the same number you will have with only using odd digits.
Two letters = 26*26 = 676 possibilities 4 digits = 104 possibilities (if numbers such as 0000 are permitted). All in all, 6.76*106 possibilities. That should keep Oxnard University going for a few years.
The significant digits in a number can be arbitrarily small or large in number, according to the method of creating them.Numbers that can have an infinite number of possible significant digits are called transcendental numbers.
There are a total of 900 3-digt numbers (100 thru 999); Of these there are 9x9x8 = 648 numbers that do not contain repeating digits (first position: 9 possibilities, second place 9 possibilities; the 10 digits minus the one used in the first place. and third place 8 possibilities; 10 digits minus the two used in the first and second places) So, there should be 900-648=252 3-digit numbers with at least one repeating digit.
For a 3 digit number, the left most or the most significant digit cannot be zero. So it can be 1,2,3,4,5,6,7,8 or 9 which is 9 possibilities. The middle number can be 0,1,2,3,4,5,6,7,8,9 which is 10 possibilties but one of the digits has been chosen already as the first digit, so the possibilities are only 9. The right most number can be 0,1,2,3,4,5,6,7,8,9 which is 10 possibilities but two of the digits have been already used by the left most and the middle digits. That leaves only 8 possibilities. So the total number of three digit numbers that have three distinct digits is 9 x 9 x 8 = 81 x 8 = 648 possibilities
There are 25 possibilities fitting those rules.
This is a permutation problem. The digits are: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The 1st digit has 9 possibilities (it cannot be zero). Thus, there are 9(9P9) = 9(9!) = 9(362,880) = 3,265,920 such numbers, because: The 2nddigit has 9 possibilities (because we are left with 9 digits). The 3rd digit has 8 possibilities. The 4th digit has 7 possibilities. The 5thdigit has 6 possibilities. The 6thdigit has 5 possibilities. The 7thdigit has 4 possibilities. The 8thdigit has 3 possibilities. The 9thdigit has 2 possibilities. The 10thdigit has 1 possibility. so that, 9*9*8*7*6*5*4*3*2*1= 3,265,920.