Question for math/stats guys

[BlancoNino]

A person going to a party was asked to bring 5 different bags of chips. Going to the store, she finds 20 varieties.

How many different selections can she make?


Yes, this is a homework question. I don't care about the answer I just want to know HOW to compute it. You can even reword it and use different numbers if you don't believe me

Thanks

Edit: Thanks Klin.

 

[KLin]

Do we have to guess what the question is in addition to doing your homework for you?

 

[PlasmaBomb]

I take it the question is how many combinations are there?

 

[BlancoNino]

Originally posted by: PlasmaBomb
I take it the question is how many combinations are there?

yes

 

[darkxshade]

Answer is 1, the amount of varieties is so great she decides to buy a bag of each.

 

[2Xtreme21]

Isn't this just a combination? nCr?

n = total
r = part

20 C 5 ? (It's been a while).

 

[PlasmaBomb]

Obligatory answer to everything - 42

 

[TecHNooB]

20!/5! is my guess. I totally don't remember how to solve these T_T

edit: nm, can't be right

 

[QED]

Originally posted by: BlancoNino
A person going to a party was asked to bring 5 different bags of chips. Going to the store, she finds 20 varieties.

How many different selections can she make?


Yes, this is a homework question. I don't care about the answer I just want to know HOW to compute it. You can even reword it and use different numbers if you don't believe me

Thanks

Edit: Thanks Klin.

Well, let's instead say she needs to select 3 different bags of chips from a selection of 10 varieties. This is a very simple combinatorics problem.

Her first selection can be from any of the 10 varieties. Her second selection can be from any of the remaining 9, and her third selection can be from any of the remaining 8. This gives a total of 10 * 9 * 8 different possibilities.

However, this is assuming that the order the chips are selected matters. That likely is not the case here, so we have to factor out all the ways that she could choose the 3 bags of chips, but in different orders. Call the 3 chips she chose A, B, C. Now she could have selected A first, second, or third, for 3 possibilities. She could have selected B from the remaining two possibilites, and C would by default have been selected at the lone remaining position. Hence, there are 3 * 2 * 1 = 6 different ways to choose 3 bags of chip with respect to order.

Hence, there are 10 * 9 * 8 / 6 = 120 different ways to choose 3 different chips from 10 varieties without respect to order.

 

[George P Burdell]

Open your textbook and read basic combinatorics.

 

[kstu]

30,517,578,125

 

[torpid]

http://www.mathsisfun.com/comb...ions-permutations.html

 

[QED]

Originally posted by: TecHNooB
20!/5! is my guess. I totally don't remember how to solve these T_T

edit: nm, can't be right

You're very close... you're missing a (20-5)! in the denominator, though.

 

[BlancoNino]

Originally posted by: QED

Originally posted by: BlancoNino
A person going to a party was asked to bring 5 different bags of chips. Going to the store, she finds 20 varieties.

How many different selections can she make?


Yes, this is a homework question. I don't care about the answer I just want to know HOW to compute it. You can even reword it and use different numbers if you don't believe me

Thanks

Edit: Thanks Klin.

Well, let's instead say she needs to select 3 different bags of chips from a selection of 10 varieties. This is a very simple combinatorics problem.

Her first selection can be from any of the 10 varieties. Her second selection can be from any of the remaining 9, and her third selection can be from any of the remaining 8. This gives a total of 10 * 9 * 8 different possibilities.

However, this is assuming that the order the chips are selected matters. That likely is not the case here, so we have to factor out all the ways that she could choose the 3 bags of chips, but in different orders. Call the 3 chips she chose A, B, C. Now she could have selected A first, second, or third, for 3 possibilities. She could have selected B from the remaining two possibilites, and C would by default have been selected at the lone remaining position. Hence, there are 3 * 2 * 1 = 6 different ways to choose 3 bags of chip with respect to order.

Hence, there are 10 * 9 * 8 / 6 = 120 different ways to choose 3 different chips from 10 varieties without respect to order.

This is exactly what I needed to know. Thanks a ton! The book's example wasn't nearly this good.

 

[PlasmaBomb]

You should have just sent a man to grab the 5 closest bags of chips...

 

[KLin]

Excel says 15504 different combinations. I think it uses a standard formulation to come up with that number. Look it up on Google.

 

[PlasmaBomb]

QED rocks

 

[PlasmaBomb]

Originally posted by: KLin
Excel says 15504 different combinations. I think it uses a standard formulation to come up with that number. Look it up on Google.

I concur.

 

[2Xtreme21]

20 C 5 = 20! / (20-5)!

Right?

 

[herm0016]

Originally posted by: 2Xtreme21
20 C 5 = 20! /(5!*(20-5)!)

Right?

fixed

 

[Schfifty Five]

Can the person choose the same bag of chips for all 5 selections?

 

[herm0016]

that would be one of the outcomes.

 

[QED]

Originally posted by: herm0016
that would be one of the outcomes.

Was that a reply to this:

Originally posted by: Schfifty Five
Can the person choose the same bag of chips for all 5 selections?

?

If so, that's not true. The answer I posed assumes all 5 chips selected are different.

The answer becomes a bit more complex if you allow duplicates in your selections.

 

[herm0016]

wait.. yea, you are right. 20C5 does not include choosing anything twice. < brain fart

 

[KLin]

Read the OP again to see if the 5 bags of chips need to be different varieties. Kthxbye.