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.


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.

  • $\begingroup$ What do you mean by the $E$s and the $R$s not coming together? Can $\cdots ER\cdots$ appear in the word? $\endgroup$ – Vee Hua Zhi Sep 13 '18 at 11:58
  • $\begingroup$ Please define the "never coming together" part more specifically. Thank you. $\endgroup$ – Vee Hua Zhi Sep 13 '18 at 11:59
  • $\begingroup$ Question Editted... $\endgroup$ – Rayees Ahmad Sep 13 '18 at 12:12
  • $\begingroup$ 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? $\endgroup$ – Vee Hua Zhi Sep 13 '18 at 12:17
  • 1
    $\begingroup$ OK... what is the answer that the book gave? $\endgroup$ – Vee Hua Zhi Sep 13 '18 at 13:02

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