3) In the word REARRANGE we have** 9 **letters.

For these letters to be distinct, we must think of each letter like a gumball in a gumball machine. Every time I use one, there of less to be used.

We also need casework for each length of word:

**CASE 1:** **1 letter words**

**9 ** one letter words.

**CASE 2: 2 letter words**

**\(9\cdot8=72\)** two letter words

**CASE 3: 3 letter words.**

\(9\cdot8\cdot7=504\)

**CASE 4: 4 letter words**

\(9\cdot8\cdot7\cdot6=3024\)

**CASE 5: 5 letter words**

**\(9\cdot8\cdot7\cdot6\cdot5=15120\)**

**CASE 6: 6 letter words**

**\(9\cdot8\cdot7\cdot6\cdot5\cdot4=60480\)**

**CASE 7: 7 letter words (almost done!)**

**\(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3=181440\)**

**CASE 8: 8 letter words**

**\(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2=362880\)**

**CASE 9: (DONE!)**

**\(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=362880\)**

Now we add all the cases together for the total number of arrangments:

\(9+72+504+3024+15120+60480+181440+2(362880)=\boxed{986409}\)

SIDENOTE:

You can subsitute the decending numbers with a **factorial**

\(5\cdot4\cdot3\cdot2\cdot1=5!\)

\(5\cdot4\cdot3=\displaystyle\frac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1}=\displaystyle\frac{5!}{2!}\)

Just some rules for future reference, hope you enjoy and use them.

As always if you have any questions just ask.

Goodnight! 🌙