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Q.
The number of even numbers greater than $100$ that can be formed by the digits $0,1,2,3$ (no digit being repeated) is
Permutations and Combinations
Solution:
The numbers are of three or four digits.
To find the number of even numbers of three digits
The unit's place must be filled by $0$ or $2$ .
$\therefore $ The number of even numbers of three digits
(having or not having $0$ in hundred's place)
$={ }^{2} P_{1} \times{ }^{3} P_{2}$.
But the number of even numbers of three digits having $0$ in hundred's place $={ }^{2} P_{1}$
($\because$ unit's place is naturally filled by $2$ and ten's place by one of $1$ and $3$ ).
$\therefore $ The number of even numbers of three digits
$={ }^{2} P_{1} \times{ }^{3} P_{2}-{ }^{2} P_{1}\,\,\,\, (1)$
Similarly, the number of even numbers of four digits
$={ }^{2} P_{1} \times{ }^{3} P_{3}-{ }^{2} P_{2} \,\,\,\,\, (2)$
Adding (1) and (2), we get the number of even numbers greater than $100$
$=\left({ }^{2} P_{1} \times{ }^{3} P_{2}-{ }^{2} P_{1}\right)+\left({ }^{2} P_{1} \times{ }^{3} P_{3}-{ }^{2} P_{2}\right)$
$=2 \times 3 \times 2-2+2 \times 3 \times 2 \times 1-2 \times 1=20$