Q.How many strings are there using the letters of the word INTERMEDIATE, if

Real arrangement problems rarely ask for a plain $n!$ or $^nP_r$ — they add a constraint (some objects repeat, or must/must-not sit together, or are confined to particular positions). Four standard techniques handle these.

1. Repeated objects (Theorem 4.4). If among $n$ objects, $p_1$ are alike of one kind, $p_2$ alike of a second kind, ..., $p_k$ alike of a $k$th kind (rest all different), the number of distinct arrangements is

since permuting the identical copies of any one kind among themselves produces no visibly new arrangement — so the naive $n!$ must be divided by $p_i!$ for every repeated kind.

2. Objects that must stay together — the string/block method. Bundle the $m$ objects that must stay adjacent into a single unit. This leaves $(n-m+1)$ units to permute ($(n-m+1)!$ ways); the $m$ bundled objects also permute among themselves inside the bundle ($m!$ ways). Total: $m!\times(n-m+1)!$.

3. Objects that must never be adjacent — the gap method. Arrange the $m=n-k$ unrestricted objects first ($m!$ ways); this creates $m+1$ gaps (including both ends). Place the $k$ restricted objects into these gaps, at most one per gap, in $^{m+1}P_k$ ways. Total: $m!\times{}^{m+1}P_k$. (The complementary count — total unrestricted permutations minus the "always together" count — is often a faster route to the never-together answer.) …