randFromArray([ new Plural(function(num) { return $.ngettext("shirt", "shirts", num); }), new Plural(function(num) { return $.ngettext("tiger", "tigers", num); }), new Plural(function(num) { return $.ngettext("book", "books", num); }), new Plural(function(num) { return $.ngettext("Fabergé egg", "Fabergé eggs", num); }), new Plural(function(num) { return $.ngettext("sock", "socks", num); }), new Plural(function(num) { return $.ngettext("action figure", "action figures", num); }) ]) randRange(5,9) randRange(3,5) factorial(NUM_THINGS) / (factorial(NUM_THINGS-NUM_TAKEN) * factorial(NUM_TAKEN))

person(1) is packing his bags for his vacation. He has `NUM_THINGS` unique plural_form(THING, NUM_THINGS), but only `NUM_TAKEN` fit in his bag.

person(1) is packing her bags for her vacation. She has `NUM_THINGS` unique plural_form(THING, NUM_THINGS), but only `NUM_TAKEN` fit in her bag.

How many different groups of `NUM_TAKEN` plural_form(THING, NUM_TAKEN) can he take?

How many different groups of `NUM_TAKEN` plural_form(THING, NUM_TAKEN) can she take?

ANSWER

person(1) has `NUM_TAKEN` spaces for his plural_form(THING), so let's fill them one by one. At first, person(1) has `NUM_THINGS` choices for what to put in the first space.

person(1) has `NUM_TAKEN` spaces for her plural_form(THING), so let's fill them one by one. At first, person(1) has `NUM_THINGS` choices for what to put in the first space.

For the second space, he only has `NUM_THINGS-1` plural_form(THING, NUM_THINGS-1) left, so there are only `NUM_THINGS-1` choices of what to put in the second space. So far, it seems like there are `NUM_THINGS \cdot NUM_THINGS-1 = NUM_THINGS * (NUM_THINGS-1)` different unique choices person(1) could have made to fill the first two spaces in his bag. But that's not quite right.

For the second space, she only has `NUM_THINGS-1` plural_form(THING, NUM_THINGS-1) left, so there are only `NUM_THINGS-1` choices of what to put in the second space. So far, it seems like there are `NUM_THINGS \cdot NUM_THINGS-1 = NUM_THINGS * (NUM_THINGS-1)` different unique choices person(1) could have made to fill the first two spaces in her bag. But that's not quite right.

Why? Because if he picked THING number 3, then THING number 1, that's the same situation as picking number 1 and then number 3. They both end up in the same bag.

Why? Because if she picked THING number 3, then THING number 1, that's the same situation as picking number 1 and then number 3. They both end up in the same bag.

So, if person(1) keeps filling the spaces in his bag, making `_.map(_.range(NUM_TAKEN), function(l){ return (NUM_THINGS - l);}).join("\\cdot") = \dfrac{NUM_THINGS!}{(NUM_THINGS-NUM_TAKEN)!} = factorial(NUM_THINGS)/factorial(NUM_THINGS-NUM_TAKEN)` decisions altogether, we've overcounted a bunch of groups.

So, if person(1) keeps filling the spaces in her bag, making `_.map(_.range(NUM_TAKEN), function(l){ return (NUM_THINGS - l);}).join("\\cdot") = \dfrac{NUM_THINGS!}{(NUM_THINGS-NUM_TAKEN)!} = factorial(NUM_THINGS)/factorial(NUM_THINGS-NUM_TAKEN)` decisions altogether, we've overcounted a bunch of groups.

How much have we overcounted? Well, for every group of `NUM_TAKEN`, we've counted them as if the order we chose them in matters, when really it doesn't. So, the number of times we've overcounted each group is the number of ways to order `NUM_TAKEN` things.

There are `NUM_TAKEN! = factorial(NUM_TAKEN)` ways of ordering `NUM_TAKEN` things, so we've counted each group of `NUM_TAKEN` plural_form(THING, NUM_TAKEN) `factorial(NUM_TAKEN)` times.

So, we have to divide the number of ways we could have filled the bag in order by number of times we've overcounted our groups.

`\dfrac{NUM_THINGS!}{(NUM_THINGS - NUM_TAKEN)!} \cdot \dfrac{1}{NUM_TAKEN!}` is the number of groups of plural_form(THING) person(1) can bring.
Another way to write this is `\binom{NUM_THINGS}{NUM_TAKEN}` , or `NUM_THINGS` choose `NUM_TAKEN`, which is `ANSWER`.

randRange(5,6) randRange(2,FRIENDS-2) factorial(FRIENDS) / (factorial(FRIENDS-SLOTS) * factorial(SLOTS))

You just got a free ticket for a boat ride, and you can bring along `SLOTS` friends! Unfortunately, you have `FRIENDS` friends who want to come along.

How many different groups of friends could you take with you?

ANSWER

There are `SLOTS` places for your friends on the boat, so let's fill those slots one by one. For the first slot, we have `FRIENDS` different choices we can make (because `FRIENDS` different friends could be in that slot).

Once we've filled the first slot, there are `FRIENDS-1` friends who could fill the second. So far, if we've filled the first two slots, and it seems like there are `FRIENDS \cdot FRIENDS-1 = FRIENDS * ( FRIENDS-1)` different choices we could have made. But that's not quite true.

Why? Because if we picked person(1), then person(2), that's the same thing as picking person(2), then person(1). They both get to be on the same boat.

So, if we continue filling the slots on our boat, making `_.map(_.range(SLOTS), function(l){ return (FRIENDS - l);}).join("\\cdot") = \dfrac{FRIENDS!}{FRIENDS-SLOTS!} = factorial(FRIENDS)/factorial(FRIENDS-SLOTS)` decisions altogether, we've overcounted a bunch of groups.

How much have we overcounted? Well, for every group of `SLOTS`, we've counted them as if the order we chose them in matters, when really it doesn't. So, the number of times we've overcounted each group is the number of ways to order `SLOTS` things.

There are `SLOTS! = factorial(SLOTS)` ways of ordering `SLOTS` things, so we've counted each group of `SLOTS` friends `factorial(SLOTS)` times.

So, we have to divide the number of ways we could have filled the boat in order by number of times we've overcounted our groups:
`\dfrac{FRIENDS!}{FRIENDS-SLOTS!} \cdot \dfrac{1}{SLOTS!}` is the number of groups we can bring on the boat trip.
Another way to write this is `\binom{FRIENDS}{SLOTS}` , or `FRIENDS` choose `SLOTS`, which is `ANSWER`.