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Q. Consider the number $N =10800$.
Number of ordered triplets ( $x, y, z)$ of positive integers such that $x y z=N$, is

Permutations and Combinations

Solution:

(ii) $ 10800=2^4 \cdot 3^3 \cdot 5^2$
Consider a, b, c as beggars and powers on 2,3 and 5 as identical objects to be distributed in a, b, c. Hence number of ways is
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NOTE: ${ }^{60} C _2$ is wrong because if we take two divisiors as $2^3 \times 3^3 \times 5^2$ and $2^4 \times 3^3 \times 5^2$ then their product will exceed the number 10800 .