Question

How many possible combinations are there if you use three characters and the characters can be letters or numbers?

Answer 1

Hint: The three characters can be either a digit (0-9) or an alphabet (a-z). So, the total possible number of characters to fill the three spaces would be \[10 + 26 = 36\]. But if we take the combination as case sensitive then we have to include capital letters (A-Z), digit (0-9) and small letters (a-z), which gives us total possible options to fill as \[10 + 26 + 26 = 62\].
Formula used:
If there are ${n_1},{n_2}...,{n_m}$ways to fill the $m$positions then the total number of ways to fill the $m$positions is given by, ${n_1} \times {n_2}. \times ... \times {n_m}$.

Complete step-by-step answer:
We can use three characters that are either an alphabet or a number to form a combination. So, there are around 26 options to use that include 10 digits and 26 alphabets. And if the letters are case-sensitive then the 26 capital letters also have to be included and the total options would be \[10 + 26 + 26 = 62\]. But, the case-sensitiveness is not mentioned in the question, so we will find the answer for both of these cases. Also, there is nothing mentioned about the repetition of the characters in the code. So, we will do both with repetition and without repetition.
Case (i): With the repetition of characters and where case-sensitiveness is ignored.
So, the first place can be filled in 36 different ways. And the second place can be filled in 36 different ways as well, as repetition is allowed. The third place also can be filled in 36 different ways. Hence, the total number of combinations is $36 \times 36 \times 36 = 46656$.

Case (ii): Without repetition of characters and where case-sensitiveness is ignored.
So, the first place can be filled in 36 different ways. And the second place can be filled in 35 different ways as the character used to fill the first position is not available now. The third place also can be filled in 34 different ways as the two characters used in the previous places are not available. Hence, the total number of combinations is$36 \times 35 \times 34 = 42840$.

Case (iii): With the repetition of characters and case-sensitiveness is considered.
So, the first place can be filled in 62 different ways. And the second place can be filled in 62 different ways as well, as repetition is allowed. The third place also can be filled in 62 different ways. Hence, the total number of combinations is $62 \times 62 \times 62 = 238328$.

Case (iv): Without repetition of characters and where case-sensitiveness is considered.
So, the first place can be filled in 62 different ways. And the second place can be filled in 61 different ways as the character used to fill the first position is not available now. The third place also can be filled in 60 different ways as the two characters used in the previous places are not available. Hence, the total number of combinations is $62 \times 61 \times 60 = 226920$.

Note: We have to understand the question well before solving these types of questions. Check whether repetition is allowed and if some other conditions are given before proceeding with the answer. Sometimes it is better to give all the possible answers if the conditions are not mentioned clearly.