Question

how many passwords of 5 characters can be made if each password must start with two digits and end with three letters? the digits will be with replacement and the letters will be without replacement.

Answer 1

Explanation:

We'Re making 5 character passwords, so i'm going to put down 5 slots. The first 2 are digits and then the last 3 are the letters the digits are with replacement the letters without okay. So how do we go about doing this? Well, i'll start with the mathematical explanation, i'm going to be intuitive so mathematical. Is that we're going to be looking for for this? It'S quite simple. There are 10 digits because it's it's without a with replacement. You just have to multiply so there's 100 different 2 digit numbers from 00 up to 99 for the letters we have a case of permutations, so it would be 26 different letters and then we're choosing 3 of them and the order matters. If the order didn't matter, it would be a combination, but the order matters here. So it's permutation- and this is equal to n factorial over n minus x, factorial, which here 26 factorial over 23 factorial, which is 26 times 25 times 24 point because of a rest, all cancels. So this becomes 16600 possibilities for 3 letter combinations. When you can't repeat a letter multiply that by this, so this is 15600 and we get a total number of possibilities as being this there's the answer. So all about was with the permutation formula. Now i'm going to do it with just an intuition okay. So how many possible letters can go here any of the 26 point? How many possible letters can go here? Well, any that you haven't already used, you can't repeat them. So there's only 25 letters left and here there's only 24 letters left and you can see you get the same answer as before:

Answer 2

Answer: 1,560,000

Step-by-step explanation: Since each password must start with two digits and end with three letters, we need to consider the number of possible combinations of digits and letters separately.

For the first two digits, there are 10 choices for each position (since we can repeat digits). Therefore, there are 10 × 10 = 100 possible combinations of digits for the first two positions.

For the last three letters, there are 26 choices for the first letter, 25 choices for the second letter (since we can't repeat the first letter), and 24 choices for the third letter (since we can't repeat the first or second letter). Therefore, there are 26 × 25 × 24 = 15,600 possible combinations of letters for the last three positions.

To find the total number of passwords, we multiply the number of possible combinations of digits by the number of possible combinations of letters:

100 × 15,600 = 1,560,000

Therefore, there are 1,560,000 possible passwords of 5 characters that can be made if each password must start with two digits and end with three letters.