If you have $n_1$ copies of word $W_1$, $n_2$ copies of word $W_2$, and so on with $n_k$ copies of word $W_k$ and $n_1+n_2+\cdots+n_k=n,$ then there are exactly
$$
\frac{n!}{n_1! n_2 ! \cdots n_k! }
$$
orderings of these words. For you, $n=24,$ and say you had 2 words repeated three times $n_1=n_2=3,$ and the rest of the words were unique, thus $n_3=\cdots=n_{20}=1.$ This number would be
$$
\frac{24!}{3!^2}
$$
which divides the original quantity by $3!^2=36$ or results in a reduction of a bit more than $5$ bits of security since $\log_2 36\approx 5$ over the 80 bits quoted in the comment to your question. See the linked notes for a full explanation.
Edit: in response to the comment below from Aman Grewal, from a discussion elsewhere it seems that the checksum is between 4 (for 12 words) and 8 (24 words) bits.
Assuming this is the case, we can just subtract 8 bits from the security parameter in bits for the version of the question here. Thus, to be concrete
$$
\mathrm{Security~ in~ bits}\approx \log_2(24!/36)-8\approx
65.86~\mathrm{bits}.
$$
The moral is don't repeat words.
https://sites.math.northwestern.edu/~mlerma/courses/cs310-05s/notes/dm-gcomb
