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 n1 objects of one kind, n2 objects of a second kind, and so on up to nk objects of a k-th kind (with n1+n2+⋯+nk≤n, the rest being distinct), then the number of distinct arrangements of all n objects is n1!n2!⋯nk!n!. 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).