A Strange Spin on Arranging Letter Combinations I was recently asked by a math nerd friend about the possible configurations of the letters ARRANGE. That is a problem which has been beat to death, but specifically, how many configurations are there if you don't need to use all the letters allowing for 6 and 5 and so on letter long words. Naturally, each letter can be used up to as many time it appears in the word ARRANGE.
I couldn't think of a good way to do this with the choose function or variations of the "must use all letters" 7!/(2!*2!).
 A: 6 letters
Not including N,G,E: $\quad3\cdot\frac{6!}{2!2!}=540$
Not including A or R: $\quad2\cdot\frac{6!}{2!}=720$
5 letters
Not including both A's or both R's: $\quad2\cdot\frac{5!}{2!}=120$
Not including 2 among N,G,E: $\quad3\cdot\frac{5!}{2!2!}=90$
Not including an A and a R: $\quad5!=120$
Not including one among N,G,E and one among A,R: $\quad3\cdot2\cdot\frac{5!}{2!}=360$
4 letters
Having only A and R: $\quad\frac{4!}{2!2!}=6$
Having 2 A's or 2 R's and 2 among R/A,N,G,E: $\quad2\cdot6\cdot\frac{4!}{2!}=144$
All 4 unique: $\quad5\cdot4!=120$
3 letters
All 3 unique: $\quad10\cdot3!=60$
Only 2 unique: $\quad2\cdot4\cdot\frac{3!}{2!}=24$
2 letters
Both same: $\quad2\cdot\frac{2!}{2!}=2$
Both unique: $\quad10\cdot2!=20$
1 letter
Number of ways: $5$
Total:
Answer $\begin{align} &= (\text{Sum of all of the above cases}) \\ &= 1260+690+270+84+22+5 \\ &= 2331 \end{align}$

Note: Numbers without "$!$" are selections and the latter part is the arrangement of those selected letters.

Since, the language of the question isn't too clear on this part so, in case, 7 letter words are also allowed then:
7 letters
Number of words: $\quad\frac{7!}{2!2!}=1260$
Total$= 3591$

Thanks to @DanielMathias for proof-reading the answer.
