How many combinations can you make with 3 characters?

Since letters can be in any combination or permutation as long as they arent words as your question suggests, any letter can occur in any of the 3 slots thus the answer is simply number of letters in the alphabet to the power of 3, also commonly known as “cubed”. Thus, 26 x 26 x 26 or 26^3 which is 17576.

How many access codes can be created that consists of a letter and two numbers?

67,600
Recall that there are 26 letters in the alphabet and there are 10 possible digits that can be chosen. Since repetition is allowed, we compute 10 · 10 · 26 · 26 = 67,600. Therefore, there are 67,600 possible ID codes consisting of 2 numbers followed by two letters when repetition of both digits and letters is allowed.

How many 6 character codes are there?

For each of the 362 ways to fill in the first two characters there are 36 ways to fill in the third character. Hence there are 363 ways to fill in the first three characters. Continuing in this fashion you can see that there are 366 = 2 176 782 336 ways to fill in all 6 characters.

How many different access codes are possible?

Originally Answered: A code consists of two letters of the alphabet is followed by 5 digits. How many such codes are possible? The answer is 67,600,000.

How many 3 character alphanumeric combinations are there?

7770 triples of distinct alphanumeric characters.

How many license plates are possible with 3 letters and 3 numbers?

Each of the three letter combinations can be combined with any of the three number combinations so the total is 17,576 x 1000 = 17,576,000 different combinations possible.

How many 3 letter and number combinations are there?

26⋅26⋅26=263=17576. If you want the letters to be unique, the calculation changes slightly.

What is the probability of selecting the correct access code on the first try?

1/1000
(a) What is the probability of randomly selecting the correct access code on the first try? (7 points) There are 10·10·10 = 1000 total options for three-digit codes. Only one is correct, so the probability of getting the correct code on the first try is 1/1000 = 0.001.