3!*3!=36
For example, if you want to use the alphabets a,b,c and three digits 1,2,3 you can make the following 36 different passwords (3 letters followed by 3 digits)
abc123
abc132
abc231
abc213
abc312
abc321
acb123
acb132
acb231
acb213
acb312
acb321
bca123
bca132
bca231
bca213
bca312
bca321
bac123
bac132
bac231
bac213
bac312
bac321
cab123
cab132
cab231
cab213
cab312
cab321
cba123
cba132
cab231
cab213
cab312
cab321
There are up to 3 numbers and up to 5 letters in a UK postcode. One or two letters, followed by one or two digits OR one digit and a letter, followed by a space, followed by one digit, followed by two letters.
2993 of them.
1757600
Using the [modern] Roman alphabet, there are 15600 possible passwords that are not case-sensitive.
If the license plate code consists of (letter-1)(letter-2)(digit-1)(digit-2)(digit-3), and the letters may not repeat, and the digits may repeat, then: There are 26 choices for (letter-1, and for each of those ... there are 25 choices for (letter-2), and for each of those ... there are 10 choices for (digit-1), and for each of those ... there are 10 choices for (digit-2), and for each of those ... there are 10 choices for (digit-3). The total number of choices is: (26 x 25 x 10 x 10 x 10) = 650 x 1,000 = 650,000
There are up to 3 numbers and up to 5 letters in a UK postcode. One or two letters, followed by one or two digits OR one digit and a letter, followed by a space, followed by one digit, followed by two letters.
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.
676,000 possibilities. For case-sensitive passwords, it's 2,704,000 possibilities. Passwords consist of 2 letters and 3 digits. That's 5 places. The first and second places can each hold 1 of 26 possibilities (26 letters), and positions 3, 4, and 5 can each hold 1 of 10 possibilities. The answer is 26 x 26 x 10 x 10 x 10 = 676000. Maybe the passwords can be case-sensitive (i.e. an upper case letter is different from a lower-case letter), then there will be 52 possibilities for each letter - in each of the 1st and 2nd places. This gives the total as 52 * 52 * 10 * 10 * 10 = 2,704,000. Note: An assumption we made is that the passwords are only 5 characters, consisting of 2 letters and 3 digits only. The question didn't say that no other characters are between these letters and numbers. If there are, then we don't have enough info to answer the question.
Passwords can be letters and /or numbers
There are 26 different letters that can be chosen for each letter. There are 10 different numbers that can be chosen for each number. Since each of the numbers/digits that can be chosen for each of the six "spots" are independent events, we can multiply these combinations using the multiplicative rule of probability.combinations = (# of different digits) * (# of different digits) * (# of different digits) * (# of different letters) * (# of different letters) * (# of different letters) = 10 * 10 * 10 * 26 * 26 * 26 = 103 * 263 = 1000 * 17576 = 17,576,000 different combinations.
Individual letters, numbers or special symbols such as @#$%^& found on a computer keyboard.
Typically one letter followed by seven numbers.
2993 of them.
1757600
Two countries beginning with the letter A contain 9 letters. The countries are Argentina and Australia.
In a "language" containing just 10 letters, there are 10,000 four-letter permutations. It's easy to work out if you simply replace the letters with decimal digits 0-9. The first permutation is 0000, followed by 0001, 0002, 0003, ..., 9997, 9998 and 9999.
the name of an identifier consists of letters and digits but name always starts with a letter.