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Q.
The total number of $3$-digit numbers, the sum of whose digits is even, is equal to
Permutations and Combinations
Solution:
Let the number be $n=p q r$.
As $p+q+r$ is even,
there are following cases:
(i) $p, q, r$ all are even $\rightarrow 4 \cdot 5 \cdot 5$
$=100$ ways
(ii) $p$ even and $q, r$ are odd $\rightarrow 4 \cdot 5 \cdot 5$
$=100$ ways
(iii) $p$ odd, $q$ and $r$ even $\rightarrow 5 \cdot 5 \cdot 5$
$=125$ ways
(iv) $p$ odd, $q$ even, $r$ odd $\rightarrow 5 \cdot 5 \cdot 5=125$
Total ways $=100+100+125+125=450$