If four fair dice are thrown simultaneously, what is the probability

If four fair dice are thrown simultaneously, what is the probability of getting at
least one pair?
\((A) \frac{1}{6}\\
(B) \frac{5}{18}\\
(C)\frac{1}{2}\\
(D) \frac{2}{3}\\
(E) \frac{13}{18}\)

One way to solve it is by considering the opposite - That on throwing the dice 4 times, you never get a pair.

In this case, the probability of what happens on the first throw does not matter. But the probability of getting a different number on the second, third and fourth throw are \(\frac{5}{6}, \frac{4}{6} and \frac{3}{6}\) respectively.

So on multiplying we get \(\frac{6}{6}*\frac{5}{6}*\frac{4}{6}*\frac{3}{6} = \frac{5}{18}\)

On subtracting it from 1, we get \(1-(\frac{5}{18}) = \frac{13}{18}\)

Therefore E. \(\frac{13}{18}\) is the correct answer

But, I wonder, Why cant we solve by the following approach?

total possible outcomes for dice 6*6*6*6 or \(6^4\)
considering the opposite outcomes or possible pairs =\(6C4=6C2=15\)

Thus, P(at least one pair)=1-\(\frac{15}{6^4}\)