Using the letters a, b, c, d, e, f find the number of 4-letter "words" such that: a) A letter can be repeated; b) Letters cannot be repeated, but can be in any order; c) Letters cannot be repeated, and must be used in alphabetical order; d) Which of the above counting question is a combination and which is a permutation? A "word" is any mixture of the letters.

Answer:When in an arrangement the order does not matter then it is Permutation,While, when the order matters then it is combination,Here, the given letters are,a, b, c, d, e, f,The number of letters in a number = 4,a) Since, each letter has 6 possible ways,Thus, if letter can be repeated, then the total number of ways = 6 × 6 × 6 × 6 = 1296,b) If letters cannot be repeated, but can be in any order,Then total number of ways = 6 × 5 × 4 × 3 = 360,c) Letters cannot be repeated, and must be used in alphabetical order,Then, the total number of ways = [tex]C(6, 4)[/tex][tex]=\frac{6!}{4!2!}[/tex][tex]=\frac{6\times 5}{2}[/tex][tex]=\frac{30}{2}[/tex][tex]=15[/tex]d) Since, in a) and b) order matters, thus they are of permutation While, in c) order does not  matter, thus it is of combination.