Anagrams of MISSISSIPPI with atleast 3 consecutive I's I'm trying to solve this question but I don't know if I am doing it right, my approach is:
Total letters (Cardinality) MISSISSIPPI: 11
Arrange 3 I's [III]ISSSSPPM
And Calculate the possible combinations
9!/(4!2!) = 7560
But I cannot verify if this is the correct solution.
 A: To count arrangements with three consecutive Is, we must arrange the nine objects [III], I, S, S, S, S, P, P, M.  To do so, we must choose four of the nine positions for the Is, two of the remaining five positions for the Ps, then arrange the three distinct objects III, I, and M in the three remaining positions.  Thus, there are
$$\binom{9}{4}\binom{5}{2}3! = \frac{9!}{4!5!} \cdot \frac{5!}{2!3!} \cdot 3! = \frac{9!}{4!2!}$$
arrangements with three consecutive Is, as you found.
However, we have counted those arrangements in which there are four consecutive Is twice, once when we designated the first three Is as the three consecutive Is and once when we designated the last three Is as the three consecutive Is.  We only want to count those cases once, so we must subtract them from the total.
If all four Is are consecutive, we have eight objects to arrange: [IIII], S, S, S, S, P, P, M.  We choose four of the eight positions for the Is, two of the remaining four positions for the Ps, then arrange the two distinct objects IIII and M in the remaining two positions.  Thus, there are
$$\binom{8}{4}\binom{4}{2}2!= \frac{8!}{4!4!} \cdot \frac{4!}{2!2!} \cdot 2! = \frac{8!}{4!2!}$$
such arrangements, as you found in the comments.
Hence, the number of anagrams of MISSISSIPPI with at least three consecutive Is is
$$\frac{9!}{4!2!} - \frac{8!}{4!2!}$$
