Till now, we calculated the permutations of objects when there were 'n' different objects taken all or 'r' at a time.

Now we will calculate the number of permutations of 'n' objects, from which 'p' of the 'n' objects are same and of one kind, 'q' of the 'n' objects are also same and of another kind, 'r' of the 'n' objects are also same and of a different kind, and the rest of the objects left are all different objects.

This is same as if there were 15 objects of different colors, and 3 were red, 3 were blue, 3 were green, and the rest 6 were all different colors.

The formula for calculating the permutations is as follows:

Let there be 'x' number of permutations when 'n' objects ('p' of one kind, 'q' of another kind, 'r' of another kind, and the rest all different) are taken all at a time.

Out of those 'n' objects, 'p' objects are same and of one kind, 'q' are also same and of another kind, 'r' objects are also same and of yet an other kind, and the rest objects are all different. If we replace the 'p' objects (that are of one kind) by 'p' number of objects of

Similarly, if we replace the 'q' objects of one kind with 'q' objects that are

Again, we replace the third set of 'r' similar objects with the same number of different objects, then the total number of permutations will increase to `x p! q! r!`.

Now we have already made all the 'n' objects different: we have replaced all the same kind of objects with different objects such that all 'n' objects are different now. We know that the total number of permutations of 'n' objects (taken all at a time) is given by `n!`. Therefore

------------------------------------------------------------------------------------------------------------

Continued >> Calculation of permutations part 6-A: Circular permutations

------------------------------------------------------------------------------------------------------------

Now we will calculate the number of permutations of 'n' objects, from which 'p' of the 'n' objects are same and of one kind, 'q' of the 'n' objects are also same and of another kind, 'r' of the 'n' objects are also same and of a different kind, and the rest of the objects left are all different objects.

This is same as if there were 15 objects of different colors, and 3 were red, 3 were blue, 3 were green, and the rest 6 were all different colors.

The formula for calculating the permutations is as follows:

`P(n) = (n!)/(p! r! q!)`

**Derivation of the above formula:**Let there be 'x' number of permutations when 'n' objects ('p' of one kind, 'q' of another kind, 'r' of another kind, and the rest all different) are taken all at a time.

Out of those 'n' objects, 'p' objects are same and of one kind, 'q' are also same and of another kind, 'r' objects are also same and of yet an other kind, and the rest objects are all different. If we replace the 'p' objects (that are of one kind) by 'p' number of objects of

**, then the number of permutations 'x' will increase because the number of different objects has increased. Since the 'p' different objects can be arranged within themselves in `p!` number of ways, this implies that the 'x' permutations will now become x * `p!` permutations.***different kinds*Similarly, if we replace the 'q' objects of one kind with 'q' objects that are

**, then the number of permutations will increase because now there are more objects that are different, and the more different objects are there, the more are the number of different arrangements that can be made from them. Since the 'q' different objects can be arranged within themselves in `q!` ways, therefore the total number of permutations will be increased to `x p! q!`.***all different*Again, we replace the third set of 'r' similar objects with the same number of different objects, then the total number of permutations will increase to `x p! q! r!`.

Now we have already made all the 'n' objects different: we have replaced all the same kind of objects with different objects such that all 'n' objects are different now. We know that the total number of permutations of 'n' objects (taken all at a time) is given by `n!`. Therefore

`n! = x p! q! r!`Therefore:

`P(n, r) = (n!)/(p! q! r!)`That is the final formula.

------------------------------------------------------------------------------------------------------------

Continued >> Calculation of permutations part 6-A: Circular permutations

------------------------------------------------------------------------------------------------------------

Thanks a lot! Best wishes! This is a wonderful work!

ReplyDelete