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We have 3 beetles (B) and 2 flies (F), total 5 insects. Number of distinct arrangements:
Step 2: Place the 2 moths (M) in non-adjacent positions.
In a sequence of 5 non-moths, there are 6 possible gaps (including ends) where moths can be inserted:
We must choose 2 distinct gaps such that they are not adjacent — but since were placing one moth per gap, any two distinct gaps ensure non-adjacency. So number of ways to choose 2 non-adjacent positions from 6 gaps is:
However, since the moths are indistinguishable, we do not multiply — each pair of gaps gives one valid placement.
Wait — correction: if two moths are placed in different gaps, even if those gaps are adjacent (e.g., between positions 2 and 3), the moths will be adjacent in the sequence unless separated by at least one non-moth — but each gap holds at most one moth.
Wait — this overcounts? Lets reconsider.
Actually, the gap method assumes that inserting one moth per gap ensures spacing. But since we are placing two moths into 6 gaps, and each gap can hold at most one, choosing any 2 of the 6 gaps gives a valid configuration with non-adjacent moths.
But this contradicts intuition — lets double-check total.
Alternative approach: total sequences with 3B, 2F, 2M.