Compute the probability that the outcome of a die’s fifth throw is one of the previous 4

Question

A die is randomly rolled five times, the probability that outcome of
the fifth throw is one of the outcomes of the first 4 throws is?

Attempt:

For first 4, there are following possibilities: (after that we will multilply the possibilities for the fifth one)

1. All different

2. 2 same 2 different

3. 3 same 1 different

4. All same

$1.$ All different:

Number of ways: $(6*5*4*3)\times\dbinom 41 = 1440$

$2$. $2$ same, $2$ different:

Number of ways: $\dfrac{4!}{2!}\dbinom 63 \times \dbinom 31 = 720$

$3$. $3$ same, $1$ different:

Number of ways: $\dfrac{4!}{3!}\dbinom 62 \times \dbinom 21 = 120$

$4$. All same:

Number of ways: $\dbinom 61 \times \dbinom 11 = 6$

Thus, $p(E) = \dfrac{@1+@2+@3+@4}{6^5} = \dfrac{381}{6^4}$

But answer given is: $\dfrac{671}{6^4}$

What's my mistake?

Edit:

Forgot to use: (as pointed out by @anryvian in comments)

$5.$ 2 same, 2 same:

Number of ways: $\dfrac{4!}{2!2!} \times \dbinom 6 2 \times 2 = 180$

Which gives the answer as: $\dfrac{411}{6^4}$ (still incorrect 🙁 )

Answer 1

An easier approach: the complementary event is "none of the first four throws is the same as the fifth throw."

$$1-\left (\frac{5}{6}\right )^4$$

I haven't yet found how to fix your attempt (which is a valid approach), but hopefully I (or someone else) catches it soon.

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Corrected computations:

1. $1440$ is correct.
2. $\frac{4!}{2!} 6 \cdot \binom{5}{2} \cdot 3 = 2160$ (You have to pick which number is the pair and which are the singles)
3. $\frac{4!}{3!} 6 \cdot 5 \cdot 2 = 240$. (You have to pick which number is the triple and which is the single)
4. $6$ is correct.
5. $180$ is correct.