randFromArray(["APE", "DEN", "CLAW", "FEAT", "IRON", "OPAL", "SONG", "RULES", "ANGER", "SLIME", "FANCY", "TIGER"]) shuffle(WORD).join("") WORD.length _.map(_.range(LENGTH), function(l){ return "_ "; }).join("") factorial(LENGTH)

How many unique ways are there to arrange the letters in the word WORD?

ANSWER

Let's try building the re-arrangements (or permutations) letter by letter. The word is `LENGTH` letters long:

BLANKS

For the first blank, we have `LENGTH` choices of letters.

After we put in the first letter, let's say it's `\text{PERM[0]}`, we have `LENGTH - 1` blanks left.

`\text{PERM[0]}` BLANKS.slice(2)

For the second blank, we only have `LENGTH - 1` choices of letters left. So far, there are `LENGTH \cdot LENGTH - 1` unique choices we can make.

We can continue in this fashion to put in a third letter, and so on. At each step, we have one fewer unique choice to make, until we get to the last letter, and there's only one we can put in.

So, the total number of unique re-arrangements must be `_.map(_.range(LENGTH), function(i){ return LENGTH - i;}).join("\\cdot")`. Another way of writing this is `LENGTH!`, or `LENGTH` factorial, which is `ANSWER`.

randRange(4, 6) ["Bloopin", "Gloopin", "Prancer", "Lancer", "Quentin", "Balthazar", "Ezekiel", "Jebediah", "Rudy"] shuffle(ALL_NAMES).slice(0, NUM_NAMES) randRange(3, NUM_NAMES - 1) factorial(NUM_NAMES) / factorial(NUM_NAMES - SLOTS)

You have `NUM_NAMES` reindeer, toSentence(NAMES), and you want to have `SLOTS` fly your sleigh. You always have your reindeer fly in a single-file line.

How many different ways can you arrange your reindeer?

ANSWER

We can build our line of reindeer one by one: there are `SLOTS` slots, and we have `NUM_NAMES` different reindeer we can put in the first slot.

Once we fill the first slot, we only have `NUM_NAMES - 1` reindeer left, so we only have `NUM_NAMES - 1` choices for the second slot. So far, there are `NUM_NAMES \cdot NUM_NAMES - 1 = NUM_NAMES * (NUM_NAMES - 1)` unique choices we can make.

We can continue in this way for the third reindeer, where we will have `NUM_NAMES - SLOTS + 1` choices.

We can continue in this way for the third reindeer, and so on, until we reach the last slot, where we will have `NUM_NAMES - SLOTS + 1` choices for the last reindeer.

So, the total number of unique choices we could make to get to an arrangement of reindeer is `_.map(_.range(SLOTS), function(l){ return (NUM_NAMES - l);}).join("\\cdot")`. Another way of writing this is `\dfrac{NUM_NAMES!}{(NUM_NAMES-SLOTS)!} = ANSWER`