A store is selling 5 types of hard candies. How many ways are there to chose?

Question

The choices are lemon, cherry, strawberry, orange, and pineapple.

How many ways are there to chose $35$ candies? I thought it would be $35^5$ because there are $35$ choices with $5$ options but that is not correct.

Next, how many ways can you get at least one of each flavor?

Then, how many ways can you get at least $2$ cherry and at least $4$ lemon?

See the Wikipedia article on stars and bars problems; it has a fairly decent explanation. — Brian M. Scott, Dec 10 '13 at 22:03
If you care about the order you choose, it would be $5^{35}$. If you don't care about the order, but just the collection you have at the end, the stars and bars answers are what you need. — Ross Millikan, Dec 10 '13 at 22:04

score 2 · Answer 1 · answered Dec 10 '13 at 21:27

You have 35 candies to choose. Let's say you have four "bars"; they separate the types of candies from each other. e.g.:

-------|-------|-------|-------|-------

Means 7 candies of each. The answer is 39C4 because this is only a rearrangement problem.

The second one, assume that each has one candies. Then it's just 30 candies to choose and 4 bars again.

Same for the last one; you choose 35-2-4=29 candies with 4 bars.

If you want to learn more, search "sticks and bars". It's a powerful method.

The name usually used here is stars and bars; the link is to the Wikipedia article, which has a decent explanation. — Brian M. Scott, Dec 10 '13 at 22:02

Daugmented · Answer 2 · 2013-12-11T10:10:02.357

Choosing $35$ candies with $5$ options, is the same as partitioning $35$ to a sum of $5$ integers (decide that every integer will be for a specific candy).
One way to do this, is looking at $35+4=39$ place holders, choosing $4$ of them as dividers and deciding that every area between placeholders represents the number of candies you take from a specific type is exactly the partition of $35$ candies to $5$ flavors.
That is, the solution to your first question is ${35+4}\choose{4}$.

The second and third follow the same logic, if you have at least one of each flavor, you have only $30$ left to choose, so the solution is ${30+4}\choose{4}$.

And finally, with $2$ cherry and $4$ lemon, there are $29$ left to choose from, so the solution is ${29+4}\choose{4}$.

Small error: with one of each flavor you have only $30$ left to choose. — Brian M. Scott, Dec 10 '13 at 22:01

score 1 · Answer 3 · edited Dec 10 '13 at 22:10

1

Answer $1$

You have $5$ types of candies, but this types are always available.

Now you want to choose $35$ candies, each time you have the possibility to choose from one of the $5$ types.

So we have $5^35$ ways, means $34359738368$ possibilities.

Answer 2

$${n\choose k} = \frac{n!}{(n-k)! k!}$$

means

${35\choose 5} = 7791168$ different ways.

edited Dec 10 '13 at 22:10

Tim Ratigan

7,247

answered Dec 10 '13 at 21:33

Erdinc Ay

133

1

I’m afraid that none of these is correct; see the answers by user43400 and Daugmented for a correct approach. – Brian M. Scott Dec 10 '13 at 22:06
thx, Mr. Scott! – Erdinc Ay Jan 19 '15 at 08:59

score 0 · Answer 4 · answered Dec 10 '13 at 21:35

I know painfully little about higher math, but perhaps I can help you phrase the question another way.

You are talking about a domain of five: l, c, s, o, p. One can think of this a 35-digit number in base 4 (we ignore the blank, or zero, value). Given that we never have an empty slot, we could count

ppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppo
pppppppppppppppppppppppppppppppppps
ppppppppppppppppppppppppppppppppppc
ppppppppppppppppppppppppppppppppppl
pppppppppppppppppppppppppppppppppop
...

So the number of possible combinations is

lllllllllllllllllllllllllllllllllll

I’m afraid that this approach doesn’t work; see the answers by user43400 and Daugmented for a correct approach. — Brian M. Scott, Dec 10 '13 at 22:08

A store is selling 5 types of hard candies. How many ways are there to chose?

4 Answers4