I've been stuck on these for a while. Please guide me through all the steps because I actually want to understand this. I've got an exam coming up.

Consider the letters in the word "MATHEMATICS". In how many ways can these 11 letters be ordered so that:
(i) The two M's are next to each other.
(ii) The two M's are next to each other but the two A's are not.


(a) Treat the two $M$'s as a single unit. So we have now 10 letters. We permute these in $\frac{10!}{2! * 2!}$ ways. Since the $M$'s are identical, we don't have to permute the order in which the two $M$'s appear.

(b) This is an inclusion exclusion problem. You have from (a) the number of ways for the two $M$'s to appear together. Now group the two $A$'s together. So there are $9!$ ways of arranging the letters so that both $M$'s and both $A$'s are together. So subtract that out from your original answer: $\frac{10!}{2! * 2!} - \frac{9!}{2!}$.

  • $\begingroup$ My answer key says something different. It says: Gluing two Ms together, we have 10 objects to arrange. There are 10!/2!2!1!1!1!1!1! such arrangements $\endgroup$ – Vimzy Apr 20 '14 at 4:42
  • $\begingroup$ I've updated my answer. Thanks for the catch! I forgot to think about multiple appearances of the same character. The two $2!$ terms in the denominator of (a) handle symmetry cases for the two $T$ and $A$ characters. The two $A$'s are glued together in (b), so we only are concerned with the $T$'s. Hence, a single $2!$ in the denominator. Sorry for the confusion. $\endgroup$ – ml0105 Apr 20 '14 at 4:44
  • $\begingroup$ Okay, that makes sense! $\endgroup$ – Vimzy Apr 20 '14 at 4:45
  • $\begingroup$ @Vimzy: There are also 2 T and 2 A. It is a multinomial permutation. $\endgroup$ – Graham Kemp Apr 20 '14 at 4:47
  • $\begingroup$ So we have two $A$ and two $T$ characters. Label them $A_{1}, A_{2}, T_{1}, T_{2}$. We divide out by $2!$ when permuting $A_{1}, A_{2}$, as permutations of the $A$ characters simply create symmetry cases. So we divide out the symmetry cases. The same applies with the $T$ characters. This is a multinomial permutation. $\endgroup$ – ml0105 Apr 20 '14 at 4:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.