(A ∪ B) = A + B - (A ∩ B)
i.e. (6 ∪ 4) = (multiples of only 6) + (multiples of only 4) - (multiples of both 4 and 6)

Multiples of only 6:
First term = \(102\)
Last Term = \(996\)
Number of terms = \(\frac{(996 - 102)}{6} + 1 = 150\)

Multiples of 4:
First term = \(100\)
Last Term = \(996\)
Number of terms = \(\frac{(996 - 100)}{4} + 1 = 225\)

But, every second number in this series is also divisible by \(8\) (with starting and ending number a multiple of 4), Therefore Multiples of only \(4 = \frac{225 + 1}{2} = 113\)

Multiples of 4 and 6:

Multiples of 4 = 100, 104, 108, 112, .....
Multiples of 6 = 102, 108, 114, 120, ...

First term common = \(108\)
Last Term common = \(996\)
Common difference would be \(24\)

Applying the formula of A.P to find the number of terms:
{NOTE: We are using this formula to find the number of terms above}

\(T_n = a + (n - 1)d\)
\(996 = 108 + (n - 1)24\)
\(n= 38\)

So, \((6 ∪ 4) = (150) + (113) - (38) = 225\)

Hence, option B