If all letters and numbers are allowed, the possibilities are 26x26x10x10x10x10. So: 6760000 different plates.
When trying to work out how many different combinations there are, you need to know how many options there are for each value. If the password only contains lower case letters, then we have 26 options for each value. For each letter in the password, there are 26 options, so the total number of possible options is 26x26x26x26x26x26 or 266 This equals 308,915,776 so there are 308,915,776 possible different combinations of six letters.
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.
I suspect that you want permutations rather than combinations. The permutation ABC is different from ACB, but they are both the same combination. There are 26*26*26*10*10 or 1,757,600 possible permutations of 3 letters followed by 2 numbers. But there are ten ways of arranging 3 letters and 2 numbers: eg LLLNN, LNLNL etc. All in all, therefore, 17,576,000 permutations. However, some letters are not used so as to avoid confusion between letters and numbers: eg 0 and O. Also, some sequences are not used because they form (or suggest) inappropriate words.
Say you have the letters A,B, and C. Here are all the possible combinations. * ABC * ACB * BAC * BCA * CAB * CBA So, 6 if you don't repeat any of the letters. If you DO repeat letters, then simply take the number of letters you have, (3 for instance), and multiply it to the power of the number of letters you have. So, for 3 letters, the formula would be 33 . Or if you had 4 letters it would be 44 and so on.
If the identifying information on each license plate consists of (letter-1)(letter-2)(digit-1)(digit-2)(digit-3)(digit-4), and repetition is allowed: letter-1 has 26 choices, and for each one ... letter-2 has 26 choices, and for each one ... digit-1 has 10 choices, and for each one ... digit-2 has 10 choices, and for each one ... digit-3 has 10 choices, and for each one ... digit-4 has 10 choices. Total number of choices = 262 x 104 = ( 26 x 26 x 10 x 10 x 10 x 10 ) = 676 x 10,000 = 6,760,000
There are 26 possible letters and 10 possible numbers. The number of license plates possible is then 26*26*10*10*10*10 = 6760000.
With no other constraints, 26 x 26 x 26 = 17,576 possible three-letter codes.
Barbara
each of the four numbers have 10 possibilities, and each of the three letters have 26 possibilities. so the total possible ways u can arrange them are 10x10x10x10x26x26x26 this totals up to 175,760,000 different license plate numbers
If you assume 26 different letters, there are 26 to the power 5 possible variations of 5 letters.
11,881,376
Consists of four numbers followed by two letters. It depends on where in Utrecht. In the city it's something between 3511 and 3585.
The word "numbers" consists of 7 distinct letters. The number of permutations of these letters is calculated using the factorial of the number of letters, which is 7!. Therefore, the total number of permutations is 7! = 5,040.
64 different arrangements are possible.
35,152,000 (assuming that 000 is a valid number, and that no letter combinations are disallowed for offensive connotations.) Also, no letters are disallowed because of possible confusion between letters and numbers eg 0 and O.
(26) x (26) x (10,000) = 6,760,000
1x26x26x26x10x10x10=17576000