How many words can be formed by arranging the letters of the word RAMANUJAN such that only one letter, i.e. N, is kept between two A's?
The solution is given as the following: [ANA]_ _ _ _ _ _ $=7!$ arrangements $-$ [ANANA]_ _ _ _ $= 5! \cdot 2$. So final answer is $4800$. Could you please elaborate?
According to the question (which is not written clearly at all), they want to find the number of arrangements where the number of N's immediately between two A's is exactly 1. So our first stab is the count the number of times where it happens at least once. That means permuting the following seven blocks of letters: $$(ANA)(R)(M)(U)(J)(A)(N)$$ which can be done in $7!=5040$ ways. The problem is that this counts arrangements where both N's are immediately between two A's, so we need to remove those arrangements to get an accurate count. Fortunately, there is only one extra A, so we need to think about the number of ways of permuting $$(ANANA)(R)(M)(U)(J)$$ But we need to be careful, because, for instance $(ANANA)(R)(M)(U)(J)$ occurred in our original count both as $(ANA)(N)(A)(R)(M)(U)(J)$ and $(A)(N)(ANA)(R)(M)(U)(J)$, so we need to subtract each of the $5!$ permutations twice to get a count. That gives us a final count of $5040-2(5!)=4800$ arrangements.