How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
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 $={ }^{5} P_{2}=\frac{5 !}{(5-2) !}=\frac{5 !}{3 !}$

$=\frac{5 \times 4 \times 3 !}{3 !}=20$

Thus, by multiplication principle, the required number of 3-digit numbers is

3 × 20 = 60

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