# 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.

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.

• What do you mean by the $E$s and the $R$s not coming together? Can $\cdots ER\cdots$ appear in the word? – Vee Hua Zhi Sep 13 '18 at 11:58
• Please define the "never coming together" part more specifically. Thank you. – Vee Hua Zhi Sep 13 '18 at 11:59
• Question Editted... – Rayees Ahmad Sep 13 '18 at 12:12
• I still do not understand what do you mean by that. Can $\cdots EE \cdots RR \cdots$ or $\cdots ER \cdots ER \cdots$ appear in the word? Or do you mean 4 of them must be separated? – Vee Hua Zhi Sep 13 '18 at 12:17
• OK... what is the answer that the book gave? – Vee Hua Zhi Sep 13 '18 at 13:02