Find the total number of permutations in which letters of the word $ARRANGEMENT$ can be permuted so that two $E$s and two $R$s do not come together.
MY TRY:
Total number of arrangements possible=$\dfrac{11!}{2!2!2!2!}=2494800$
Number of arrangements that are possible from two $E$s and two $R$s =$\dfrac{4!}{2!2!}=6$
Now if we treat two $E$s and two $R$s as one unit then number of possible arrangements=$\dfrac{6.8!}{2!2!}=60480$
Total number of required permutations$= 2494800-60480=2434320$
But the answer provided by the text does not match my answer. Please correct me.