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Question
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Solution
3-digit even numbers are to be formed using the given six digits, 1, 2, 3, 4, 6, and 7, without repeating the digits.
Then, units digits can be filled in 3 ways by any of the digits, 2, 4, or 6.
Since the digits cannot be repeated in the 3-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining 5 digits.
Therefore, the number of ways in which hundreds and tens place can be filled with the remaining 5 digits is the permutation of 5 different digits taken 2 at a time.
Number of ways of filling hundreds and tens place
= 5P2 = `(5!)/((5 - 2)!) = (5!)/(3!)`
= `(5 xx 4 xx 3!)/(3!)`
= 20
Thus, by multiplication principle, the required number of 3-digit numbers is 3 × 20 = 60
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