Calculating Five-Digit Numbers with a Repeated Digit

The problem asks for the number of five-digit numbers that can be formed using the digits {3, 4, 5, 6} such that exactly one digit appears twice. This means the five digits used will consist of 4 distinct digits from the set {3, 4, 5, 6}, with one of these 4 digits repeated.

Step 1: Identify the structure of the digits

A five-digit number is required. The digits must be chosen from the set {3, 4, 5, 6}. Exactly one digit must appear twice. This implies the remaining $5 - 2 = 3$ digits must be distinct and different from the repeated digit. Therefore, the number must be formed using 4 distinct digits in total.

Step 2: Select the digits

• Choose the 4 distinct digits: Since we must use digits from the set {3, 4, 5, 6}, and we need 4 distinct digits, we must choose all of them. The number of ways to choose 4 digits from {3, 4, 5, 6} is $\binom{4}{4} = 1$. The set of distinct digits is {3, 4, 5, 6}.
• Choose the digit to repeat: From these 4 distinct digits, we need to select one digit that will appear twice in the five-digit number. The number of ways to choose this digit is $\binom{4}{1} = 4$.

Let the chosen repeated digit be $X$. The other three distinct digits are $Y, Z, W$. The set of five digits to be arranged is {$X, X, Y, Z, W$}.

Step 3: Arrange the digits

Now, we need to find the number of distinct permutations of these 5 digits {$X, X, Y, Z, W$}. This is a permutation problem with repetitions. The formula for permutations of n objects where there are $n_1$ identical objects of type 1, $n_2$ identical objects of type 2, ..., $n_k$ identical objects of type k is $\frac{n!}{n_1! n_2! ... n_k!}$. In our case, $n = 5$ (total number of digits). One digit ($X$) appears twice, so $n_1 = 2$. The other three digits ($Y, Z, W$) appear once each, so $n_2 = 1, n_3 = 1, n_4 = 1$. The number of arrangements is: $ \frac{5!}{2! \times 1! \times 1! \times 1!} = \frac{5!}{2!} $ Calculating the factorial values: $ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $ $ 2! = 2 \times 1 = 2 $ So, the number of permutations is: $ \frac{120}{2} = 60 $ There are 60 distinct ways to arrange the five digits once the set of digits {$X, X, Y, Z, W$} is determined.

Step 4: Calculate the total count

To find the total number of five-digit numbers, we multiply the number of ways to choose the digits by the number of ways to arrange them. Total numbers = (Ways to choose the repeated digit) $\times$ (Ways to arrange the 5 digits) Total numbers = $4 \times 60$ $ \text{Total numbers} = 240 $

Therefore, 240 five-digit numbers can be formed using the integers 3, 4, 5, and 6 with exactly one digit appearing twice.