Correct Option (B)

The problem requires determining the maximum exponent 'm' of 3 in the prime factorization of 15!. This is equivalent to finding the total number of times the prime factor 3 appears in the product $1 \times 2 \times 3 \times . . . \times 15$ .

To find the exponent of a prime 'p' in $n!$ , the standard method involves summing the floor values of $n$ divided by successive powers of $p$ :

$E_{p} (n!) = ⌊ \frac{n}{p} ⌋ + ⌊ \frac{n}{p^{2}} ⌋ + ⌊ \frac{n}{p^{3}} ⌋ + . . .$

In this specific case, $n = 15$ and $p = 3$. We calculate the terms as follows:

First term: $\left\lfloor \frac{15}{3} \right\rfloor = 5$. This counts the multiples of 3 (3, 6, 9, 12, 15), each contributing at least one factor of 3.
Second term: $\left\lfloor \frac{15}{3^2} \right\rfloor = \left\lfloor \frac{15}{9} \right\rfloor = 1$. This accounts for multiples of 9 (only 9 within the range 1 to 15), which contribute an additional factor of 3 (since 9 = 3 x 3, one factor was already counted in the first term).
Third term: $\left\lfloor \frac{15}{3^3} \right\rfloor = \left\lfloor \frac{15}{27} \right\rfloor = 0$. Higher powers of 3 will also yield 0, as $3^k > 15$ for $k \ge 3$.

Summing these values: $5 + 1 + 0 = 6$.

Therefore, the maximum value of m is 6.

Incorrect Options:

Option (A) 7: This value is incorrect. It may result from an arithmetic error or a misapplication of the method for counting prime factors in a factorial.
Option (C) 5: This value is incorrect. It would be obtained if only the first term $⌊ \frac{15}{3} ⌋$ was considered, thereby omitting the additional factor of 3 contributed by multiples of 9 (e.g., 9 contributes two factors of 3, but only one would be accounted for in the first term).
Option (D) 4: This value is incorrect and represents a significant underestimation of the number of prime factors of 3 in 15!.