Explanation:

We will use three key concepts to solve this question;

1. When we divide \(25!\) by \(10^7\), the last 7 digits would be the remainder
ex: if we divide \(5!\) by 100, the last 2 digits would be the remainder
\(5! = (5)(4)(3)(2)(1) = 120\)
Remainder = 20

2. In order to find the maximum power of any number (say \(a\)) in \(N!\) (say), divide \(N!\) by the increasing power of \(a\) till the numerator remains greater than the denominator
ex: to find number of 2's in \(4!\):
\(\frac{4}{2^1} + \frac{4}{2^2}\)
\(2 + 1 = 3\)
Therefore, number of 2's in \(4!\) is 3

3. Cyclicity of 2, 3, 7 and 8 is 4 whereas cyclity of 4 and 9 is 2
i.e. the units digit of 2, 3, 7, and 8 repeat every 4 powers and units digit of 4 and 9 repeat every 2 powers

Coming back to the Question:

Find the number of 2's, 3's, 5's, 7's, 11's, 13's, 17's, 19's and 23's in \(25!\) by using the second concept.

Number of 2's = 22
Number of 3's = 10
Number of 5's = 6
Number of 7's = 3
Number of 11's = 2
Number of 13's = 1
Number of 17's = 1
Number of 19's = 1
Number of 23's = 1

i.e. \(25! = (2^{22})(3^{10})(5^6)(7^3)(11^2)(13)(17)(19)(23)\)

\(25! = (2^{16})(3^{10})(7^3)(11^2)(13)(17)(19)(23)(10^6)\)

Now, Units digit of:
\((2^{16})\) is 6
\((3^{10})\) is 9
\((7^3)\) is 3
\((11^2)\) is 1
\((13)\) is 3
\((17)\) is 7
\((19)\) is 9
\((23)\) is 3

So, \(25! = (ab6)(cd9)(ef3)(gh1)(ij3)(kl7)(mn9)(op3)(10^6) = abcdefghijklmnop 4(10^6)\)

Col. A: \(4(10^6)\)
Col. B: \(2(10^6)\)

Hence, option A