Permutations with Repetition:

(a) The number of permutations (arrangements) of n different type of objects, taken r at a time, when each and every object can appear once, twice, thrice.... upto r times in some arrangement

            = The number of types of filling r places where every place may be substituted by some one of n objects .

The number of permutations = The number of types of filling r places = (n)^r

(b) The various number of arrangements that will be formed using n objects out of which p are identical (and of one kind), q are identical (and of another kind), r are identical (and of another kind) and the rest are distinct is .

Example:        How many 3 digit numbers may be prepared using the digits 0, 1,2,3,4,5 so that

                              (a)       digits cannot be repeated     

                              (b)       digits can be repeated

Solution:                  Suppose the 3-digit number be XYZ

                                          Position (X) may be fulfilled by 1,2,3,4,5 but not 0. So it may be fulfilled in 5 types.

                                          Position (Y) may be filled in 5 types again. (Since 0 can be placed in this postion).

                                          Position (Z) can be filled in 4 ways.

                                          Hence, by the fundamental rule of counting, total number of types is

                                          5 x 5 x 4 = 100 ways.

                              (b)       Suppose the 3 digit number be XYZ

                                          Position (X) may be filled in 5 types

                                          Position (Y) may be filled in 6 types.

                                          Position (Z) may be filled in 6 types.

                                          Therefore by the fundamental rule of counting, total number of types is

                                          5 x 6 x 6 = 180.

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