# How many factors of \[{2^5} \times {3^6} \times {5^2}\] are perfect squares?

A. 24

B. 12

C. 16

D. 22

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**Hint:**First of all, find the possible number of ways in which perfect square factors of \[{2^5},{3^6},{5^2}\] can be arranged individually. Then use the multiplicative principle of permutations to get the required answer. So, use this concept to reach the solution of the given problem.

__Complete step-by-step answer__:For a perfect square, the power of each should be even.

The possible factors of \[{2^5}\] are \[{2^0},{2^1},{2^2},{2^3},{2^4},{2^5}\]

So, the possible perfect square factors of \[{2^5}\] are \[{2^0},{2^2},{2^4}\].

Therefore, possible number of ways of arranging the perfect square factors of \[{2^5}\] = 3

The possible factors of \[{3^6}\] are \[{3^0},{3^1},{3^2},{3^3},{3^4},{3^5},{3^6}\]

So, the possible perfect square factors of \[{3^6}\] are \[{3^0},{3^2},{3^4},{3^6}\].

Therefore, possible number of ways of arranging the perfect square factors of \[{3^6}\] = 4

The possible factors of \[{5^2}\] are \[{5^0},{5^1},{5^2}\]

So, the possible perfect square factors of \[{5^2}\] are \[{5^0},{5^2}\].

Therefore, possible number of ways of arranging the perfect square factors of \[{5^2}\] = 2

By using multiplicative principle of permutations, we have

The total number of ways of arranging the perfect square factors of \[{2^5} \times {3^6} \times {5^2}\] are \[3 \times 4 \times 2 = 24\]

Hence there are 24 factors of \[{2^5} \times {3^6} \times {5^2}\] which are perfect squares.

**Thus, the correct option is A. 24**

**Note**: In this problem we have used multiplicative principle permutations i.e., if there are \[x\] number of ways of arranging one thing and\[y\] number of ways of arranging another, then the total number of ways of arranging both the things is given in \[xy\] number of ways.