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