Permutation Of Objects When Few Are Identical - Practice Questions & MCQ

If there n objects of which p objects are of one type, q objects of another type, r objects of yet another type, and all others are distinct, the total number of ways of arranging all the objects is $\frac{\mathrm{n}!}{\mathrm{p}!\mathrm{q}!\mathrm{r}!}$

Proof: 

Suppose the total number of permutations is x, now if we replace all p identical objects by p different objects then we have x × p! arrangements. The number of arrangements, if we do the same thing with q and r, will be x × p!×q!×r!

Now, we have replaced all identical objects and we are left with n different object which can be arranged in n! Ways.

Hence, x × p!×q!×r! = n!

So, $\mathrm{x}=\frac{\mathrm{n}!}{\mathrm{p}!\mathrm{q}!\mathrm{r}!}$

Example: In how many ways can the letters of the word “MISSISSIPPI” be arranged?

Solution: repeated letters I = 4 times, S = 4 times and P=2 times

So using the above formula we have $x=\frac{11!}{4!\times 4!\times 2!}$