how many ways can the letters in ARRANGEMENT can be arranged Using all the letters of the word ARRANGEMENT how many different words using all letters at a time can be made such that both A, both E, both R both N occur together .
 A: The word ARRANGEMENT has $11$ letters, not all of them distinct. Imagine that they are written on little Scrabble squares. And suppose we have $11$ consecutive slots into which to put these squares. 
There are $\dbinom{11}{2}$ ways to choose the slots where the two A's will go. For each of these ways, there are $\dbinom{9}{2}$ ways to decide where the two R's will go. For every decision about the A's and R's, there are $\dbinom{7}{2}$ ways to decide where the N's will go. Similarly, there are now $\dbinom{5}{2}$ ways to decide where the E's will go. That leaves $3$ gaps, and $3$ singleton letters, which can be arranged in $3!$ ways, for a total of 
$$\binom{11}{2}\binom{9}{2}\binom{7}{2}\binom{5}{2}3!.$$
A: "ARRANGEMENT" is an eleven-letter word.
If there were no repeating letters, the answer would simply be $11!=39916800$.
However, since there are repeating letters, we have to divide to remove the duplicates accordingly.
There are 2 As, 2 Rs, 2 Ns, 2 Es
Therefore, there are $\frac{11!}{2!\cdot2!\cdot2!\cdot2!}=2494800$ ways of arranging it.
