【Example 9】Fired 10 shots, it is known: (1) Hit 7 shots, including one "5 consecutive hits", the remaining 2 shots are not consecutive; (2) Only one "4 consecutive hits" formed by consecutive hits is known, other situations are unclear; (3) Hit 6 shots, including one "4 consecutive hits", it is unknown whether the remaining 2 shots are consecutive. How many different possible scenarios are there in each of the above three situations?
Solution (1) There are 3 misses, and the "5 consecutive hits" and the other 2 non-consecutive hits should be distributed in the different intervals formed by them, and there are 4 such intervals:
$$
\text { (1) } \times \text { (2) } \times \text { (3) } \times \text { (4) }
$$

Therefore, it involves multiple combinations of 4 intervals: one interval for the 5 consecutive hits, and the other two intervals each for one hit, so there are $\binom{4}{12}=\frac{4!}{1!2!1!}=$ 12 different cases.
(2) The "4 consecutive hits" can appear at the beginning, or at the end, or in the middle. When the 4 consecutive hits appear in the first 4 shots, the 5th shot must be a miss (otherwise, it would be more than 4 consecutive hits). When it appears in the last 4 shots, the 6th shot must be a miss. In these two cases, the remaining 5 shots can either all hit or all miss, with $2^{5}$ different possible situations each. When the 4 consecutive hits appear in the middle, the shots before and after must both be misses, and the remaining 4 shots can either hit or miss, with $2^{4}$ different possible situations. However, the specific position where the 4 consecutive hits appear has 5 different possibilities: shots 2 to 5, shots 3 to 6, shots 4 to 7, shots 5 to 8, and shots 6 to 9. Therefore, there are $2^{5}+2^{5}+5 \times 2^{4}=$ 144 different possible situations.
(3) It should be divided into two scenarios: the other 2 shots are not consecutive and the other 2 $\frac{5!}{1!2!2!}+\frac{5!}{1!1!3!}=50$ different possible situations.

From the above examples, we have seen that as a special case of the multi-group combination model, the permutation of not entirely distinct elements has a broad application background and is a counting model that should be studied.