Forming Two-Digit Numbers Without Repetition

The problem asks us to find the total number of distinct two-digit numbers that can be formed using the digits {1, 2, 3, 4, 5, 6} with the condition that the digits are not repeated.

Calculating the Number of Two-Digit Numbers

We have 6 distinct digits available: {1, 2, 3, 4, 5, 6}. We need to form two-digit numbers.

• Tens Place: We can choose any of the 6 available digits for the tens place. So, there are 6 options.
• Units Place: Since repetition is not allowed, after choosing a digit for the tens place, we have 5 remaining digits left. We can choose any of these 5 digits for the units place. So, there are 5 options.
• Total Numbers: To find the total number of two-digit numbers possible without repetition, we multiply the number of options for each place. Total Numbers = (Options for Tens Place) $\times$ (Options for Units Place) Total Numbers = $6 \times 5$ Total Numbers = 30

Alternatively, this is a permutation problem where we need to arrange 2 digits out of 6. The formula for permutations is $P(n, k) = \frac{n!}{(n-k)!}$. Here, $n=6$ (total digits) and $k=2$ (digits to choose). $P(6, 2) = \frac{6!}{(6-2)!} = \frac{6!}{4!} = \frac{6 \times 5 \times 4!}{4!} = 6 \times 5 = 30$.

Therefore, there are 30 unique two-digit numbers that can be formed from the digits 1, 2, 3, 4, 5, and 6 without repetition.