Given,

The word ARRANGE. 

It has 7 letters of which 2 letters (A, R) are repeating. The letter A is repeated twice, and the letter R is also repeated twice in the given word. 

To find : Number of ways the letters of word ARRANGE be arranged in such a way that not all R’s do come together.

First, 

We find all arrangements of word ARRANGE, and then we minus all those arrangements of word ARRANGE in such a way that all R’s do come together, from it. 

This will exactly be the same as- all number of arrangements such that not all R’s do come together.

Since we know, 

Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!.

And, 

We also know, 

Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is \(\frac{n!}{p!\times q!\times r!}\).

i.e., 

The number of repeated objects of same type are in denominator multiplication with factorial.

A total number of arrangements of word ARRANGE : 

Total letters 7. 

Repeating letters A and R, 

The letter A repeating twice and letter R repeating twice. 

The total number of arrangements.

= \(\frac{7!}{2!\times 2!}\)

Now,

We find a total number of arrangements such that all R’s do come together.

A specific method is usually used for solving such type of problems. 

According to that, 

We assume the group of letters that remain together (here R, R) is assumed to be a single letter and all other letters are as usual counted as a single letter. 

Now,

Find a number of ways as usual; 

The number of ways of arranging r letters from a group of n letters is equals to ⁿP_r. 

And the final answer is then multiplied by a number of ways we can arrange the letters in that group which has to be stuck together in it (Here R, R).

Letters in word ARRANGE : 7 letters 

Letters in a new word : 

A, RR, A, N, G, E : 6 letters (Letter A repeated twice).

Total number of word arranging all the letters \(\frac{6!}{2!}\times\)\(\frac{2!}{2!}\)

Where second fraction 2! divided by 2! comes from arranging letters inside the group RR : 

Arrangements of two letters where all the two letters are same = 2!/2! = 1 (Obviously! You can even think of it).

Now, 

A Total number of arrangements where not all R’s do come together is equals to total arrangements of word ARRANGE minus the total number of arrangements in such a way that all R’s do come together.

= \(\frac{7!}{2!\times 2!}\) - (\(\frac{6!}{2!}\times\)\(\frac{2!}{2!}\)) 

= 900 

Hence, 

The total number of arrangements of word ARRANGE in such a way that not all R’s come together is equals to 900.