Forming 5-Digit Numbers Divisible by 5

This solution determines the count of 5-digit numbers formed from the digits {0, 2, 3, 4, 6, 7, 9} without repetition, which are divisible by 5.

Key Constraints and Conditions

• The numbers must have 5 digits.
• Available digits: {0, 2, 3, 4, 6, 7, 9} (total 7 digits).
• Each digit can be used at most once.
• The first digit cannot be 0 (to ensure it's a 5-digit number).
• The number must be divisible by 5.

Divisibility by 5 Rule

For a number to be divisible by 5, its last digit must be either 0 or 5.

Analyzing Digit Availability

In the given set {0, 2, 3, 4, 6, 7, 9}, the digit 5 is not present. Therefore, the only possibility for the last digit is 0.

Calculating the Number of Combinations (Case: Last Digit is 0)

Consider the 5 positions for the digits: $\text{P}_1 \text{P}_2 \text{P}_3 \text{P}_4 \text{P}_5$.

• Step 1: Unit's Place ($\text{P}_5$) The last digit ($\text{P}_5$) must be 0. There is only 1 way to choose this digit. $ \text{Number} = \text{_ _ _ _ 0} $
• Step 2: Remaining Digits and Positions After placing 0 at the end, we have 6 remaining digits: {2, 3, 4, 6, 7, 9}. We need to fill the first 4 positions ($\text{P}_1, \text{P}_2, \text{P}_3, \text{P}_4$) using these 6 digits without repetition.
• Step 3: Permutations for the First Four Positions The number of ways to arrange 4 distinct digits chosen from a set of 6 is calculated using permutations. The formula is $P(n, k) = \frac{n!}{(n-k)!}$. Here, $n=6$ (remaining digits) and $k=4$ (positions to fill). $ P(6, 4) = \frac{6!}{(6-4)!} = \frac{6!}{2!} $
• Step 4: Calculation $ P(6, 4) = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 6 \times 5 \times 4 \times 3 = 360 $ Since the first digit ($\text{P}_1$) cannot be 0 (and 0 is already used in $\text{P}_5$), this calculation correctly gives the number of valid 5-digit numbers.

Final Count

There are 360 possible 5-digit numbers that meet the given conditions.