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Q. Total number of 5 digit numbers having all different digits and divisible by 4 that can be formed using the digits $\{1,3,2,6,8,9\}$, is equal to

Permutations and Combinations

Solution:

A number is divisible by four, if the last two digit are divisible by four. In this case last two digits can be $12, 16, 28, 32, 36, 68, 92$ or $96$.
Total number of such numbers $=8\left({ }^4 C _3 \cdot 3 !\right)=192$