A problem in Permutation Q. If all the letters of word ARRANGE are arranger in all possible ways,in how many of words we will have the A's not together and also R's not together?
Attempt:
Required no. of permutations=No. of ways ARRANGE can be rearranged-No of ways ARRANGE can be rearranged with A's & R's together.
=$7!/(2!*2!)-5!$
But,it doesnot give the answer.
Where is my mistake?

Use inclusion/exclusion principle:


*

*Include the total number of words: $\binom72\cdot\binom52\cdot\binom31\cdot\binom21\cdot\binom11=1260$

*Exclude the number of words with AA: $\binom62\cdot\binom41\cdot\binom31\cdot\binom21\cdot\binom11=360$

*Exclude the number of words with RR: $\binom62\cdot\binom41\cdot\binom31\cdot\binom21\cdot\binom11=360$

*Include the number of words with AA and RR: $\binom51\cdot\binom41\cdot\binom31\cdot\binom21\cdot\binom11=120$


Hence the number of words with no AA and no RR is $1260-360-360+120=660$.
