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

$\begin{array}{1 1}(A)\;70\\(B)\;80\\(C)\;90\\(D)\;60\end{array}$

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• $nP_r=\large\frac{n!}{(n-r)!}$
Let 2 be fixed at unit's place.Now we have 5 digits and 3 places are to be filled up.
This can be alone in $5P_2$ ways.
When unit's place is filled up both 4 or 6 again in each case we have $5P_3$ numbers
$\therefore$ 4-digit even numbers are
$3\times 5P_2=3\times \large\frac{5!}{(5-2)!}$
$\Rightarrow 3\times \large\frac{5!}{3!}$
$\Rightarrow 3\times \large\frac{5\times 4\times 3\times 2\times 1}{3\times 2\times 1}$
$\Rightarrow 3\times 5\times 4$
$\Rightarrow 60$
Hence (D) is the correct answer.