How many 5-digit multiples of 5 can be formed from the digits 2, 3, 5, 7, 9 when no digit is repeated?
How many 5-digit multiples of 5 can be formed from the digits 2, 3, 5, 7, 9 when no digit is repeated?
Explanation
Step 1: Rule for divisibility by 5
A number is divisible by 5 if and only if it ends in 0 or 5.
But 0 is not among the given digits, so the last digit must be 5.
Step 2: Fix the last digit as 5
So, we fix 5 as the last digit.
Now we must choose the remaining 4 digits (from the remaining digits: 2, 3, 7, 9)
These will go in the first 4 places.
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Number of ways to arrange 4 digits out of 4 = 4! = 24