To use a certain cash machine, you ned a persona identification code. If each code consists of two letters followed by one of the digits from 1-9(such as AQ7 or BB3) or one letter followed by two difits (such as Q3 or J88), how many different codes can be assigned?
I think there are 9 numbers but my teacher said I have to use 10?
Oh ok. I think numbers and letters can be repeated, but thank you for clearing up the amount of numbers.
To calculate the total number of different codes that can be assigned, you need to consider the possibilities for each component of the code separately.
For the two-letter component, there are 26 letters in the alphabet, so the number of possibilities would be 26 * 26 = 676.
For the three-digit component, there are two cases:
1. The first digit is a letter and the next two digits are numbers: There are 26 possibilities for the letter and 9 possibilities for each of the digits. So, the number of possibilities would be 26 * 9 * 9 = 2,106.
2. The first digit is a number and the next two digits are also numbers. In this case, there are 10 possibilities for each of the digits. So, the number of possibilities would be 10 * 10 * 9 = 900.
To find the total number of different codes, you sum up the possibilities from each case: 676 + 2,106 + 900 = 3,682.
Therefore, there are 3,682 different codes that can be assigned using this format.
To calculate the number of different codes that can be assigned, we need to consider the different options for each position in the code.
For the first position, we can have either a letter or a digit. Since there are 26 letters in the alphabet, we have 26 options for the first letter. Additionally, we can have any of the 9 digits from 1 to 9, so we have a total of 26 + 9 options for the first position.
For the second position, we can also have either a letter or a digit. So, we have the same number of options as before - 26 + 9 options.
For the third position, we can only have a digit from 1 to 9. Therefore, we have 9 options for the third position.
To calculate the total number of different codes, we can multiply the number of options for each position:
(26 + 9) * (26 + 9) * 9 = 35 * 35 * 9 = 11,025.
So, there are a total of 11,025 different codes that can be assigned.
Regarding your confusion about using 10 numbers, it might be possible that your teacher is considering the number 0 as an additional digit. In this case, you would have 10 digits from 0 to 9 instead of 9 digits from 1 to 9. If this is the case, you would have:
(26 + 9) * (26 + 9) * 10 = 35 * 35 * 10 = 12,250 different codes.
Well, there are 9 digits from 1-9, not 10, that would include the 0
I will also assume that letters and numbers may be repeated.
case1 , 2 letters followed by a number
number of ways = 26x26x9 = 6048
case 2, 1 letter followed by 2 numbers
number of ways = 26x9x9 = 2106
total = 8154
of no digits or letters may be repeated, the number of ways would be
26x25x9 + 9x8x26
= ....