Q.In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all person of the same nationality sit together?
A) 3! 8! B) 3! 4! 8! 4! C) 4! 4! D) 8! 4! 4!

When the $n$ objects to be arranged are not all distinct — that is, some objects are exact, indistinguishable copies of each other — the plain permutation count $n!$ overcounts, because swapping two identical copies with each other produces an arrangement that looks exactly the same, yet $n!$ would have counted it as if it were different. The correction is to divide out the internal rearrangements of each group of identical objects: if a set of $n$ objects contains $n_1$ objects of one kind, $n_2$ objects of a second kind, and so on up to $n_k$ objects of a $k$-th kind (with $n_1+n_2+\cdots+n_k\le n$, the rest being distinct), then the number of distinct arrangements of all $n$ objects is $\dfrac{n!}{n_1!\,n_2!\,\cdots\,n_k!}$. This is exactly the formula used to count the distinct arrangements of the letters of a word with repeated letters (like arranging the letters of BANANA, where A repeats 3 times and N repeats twice), or of any collection with repeated items (a shelf of books that are identical within each subject, marbles of only a few distinct colours, or a sequence of coin-toss outcomes). A frequent companion technique is the 'bundle a required group together as one block, then arrange the block among the other units, then arrange within the block' method for handling 'these specific objects must/must not be adjacent' conditions, often combined with inclusion–exclusion when TWO separate adjacency conditions are imposed at once (e.g. neither of two repeated-letter pairs may be adjacent).