Question

How many different two-digit numbers can be formed from the digits 3, 1, 4, and 5 (allowing reuse)?

Answer 1

We have been given 4 digits from which multiple 2-digit terms are to be formed. Also, the reuse of the digits is allowed. This implies that we have to find the total number of permutations of the 2-digit numbers which are formed by rearranging the four given digits. Thus, we shall use the formula of permutations to calculate the factorial of total number of options available for each of the two digits.

Complete step-by-step answer:
Given the digits, 3, 1, 4, and 5.
In order to form a 2-digit number, we have to choose digits for both the digits.
Two digit number: $\underline{{{n}_{c}}}\text{ }\underline{{{n}_{c}}}$
We shall first find the total number of choices we have to fill each of the two digits of the 2-digit number. We have been given 4 digits and since the reuse of these 4 digits is allowed in the 2-digit number formed, thus the total number of choices for each digit is 4.
$\Rightarrow {{n}_{c}}=1$
Two digit number: $\underline{4}\times \underline{4}$
Hence, we get that the total number of numbers that can be formed are ${{2}^{4}}$.
We know that ${{2}^{4}}=2\times 2\times 2\times 2$
$\Rightarrow {{2}^{4}}=16$
Therefore, the total number of different two-digit numbers that can be formed from the digits 3, 1, 4, and 5 (allowing reuse) is 16.

Note: If the reuse of digits would not have been allowed to form the 2-digit number, then the total number of choices for the first digit would be equal to 4 and the total number of choices for the second digit would be equal to 3. Therefore, the total number of digits formed would be equal to $\underline{4}\times \underline{3}$.