Question
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Answer

We know that, even number means that the last digit should be even,
No. of possible digits at one’s place = 3 (2, 4 and 6)
$\Rightarrow$ No. of permutations = $_{1}^{3} \mathrm{P}=\frac{3 !}{(3-1) !}$ = 3
One of a digit is taken at one’s place, Number of possible digits available = 5
$\Rightarrow$ No. of permutations = $_{2}^{5} P=\frac{5 !}{(5-2) !}=\frac{5 \times 4 \times 3 !}{3 !}$ = 20
Thus, a total number of permutations = 3 $\times$ 20 = 60

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