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tstockero · Feb 3, 2009, 07:02 PM

My daughter needs help with the following type of questions: I can't figure out the formula.
If there are 20 different cars, how many 3 combo types can you make?:confused:

bones252100 · Feb 3, 2009, 08:30 PM

With no other info than the 2 numbers the answer is 3600. With more info, someone on this helpdesk can certainly & will want to help. Repost with more info.

tstockero · Feb 3, 2009, 08:42 PM

The exact question is: "If you can choose from a fleet of 20 train cars, how many ways can you arrange a 3 car train? (any car can be in the front).
What is the formula for figuring this out?

galactus · Feb 3, 2009, 08:59 PM

Assuming order does not matter, the formula is

$C(20,3)=1140$ ways.

tstockero · Feb 3, 2009, 10:03 PM

OK, I still don't understand. Can you make it even simpler for those of us who majored in stuff other than math. I can't believe this is my 10 year olds homework and I'm confused. Thanks for your help with this... :eek:

Unknown008 · Feb 4, 2009, 03:42 AM

Gosh! I learned that a year ago!

The way I learned it was, choosing 3 out of 20 was done with the formula

$^{20}C_{3}$ or simplified

$\frac{{20}\times{19}\times{18}}{1\times2\times3}$

galactus · Feb 4, 2009, 08:36 AM

Why is it every time I try to rate an answer I get this:

"you must spread some Reputation around before giving it to ****** again."

Unknown is certainly correct. It is based on the formula for combinations:

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

In your case, n=20 and k=3

If you don't know what n! is that means 'factorial'. That exclamation point stands for factorial not because we are excited about it.:p

A factorial is multiplying the number by each consecutive number down to 1.

For instance, 5! Would be 5*4*3*2*1=120.

So, you can see that 20! Is pretty big.

So, we have:

$\frac{20\cdot 19\cdot 18\cdot\cdot\cdot 1}{3\cdot 2\cdot 1 (17\cdot 16\cdot\cdot 15\cdot\cdot\cdot 1)}$

Now, we can see that the 17*16*15... cancels with the same in the numerator and we are left with:

$\frac{20\cdot 19\cdot 18}{6}=1140$

Just as unknown showed.

Does that help any understanding combinations?

ebaines · Feb 4, 2009, 10:54 AM

For a ten year old's homework this has to be solved without knowing anything about factorials, permutations, combinations, etc. The child is not in high school! So approach the problem logically, step by step:

Suppose the 20 train cars are labelled A, B, C, etc up to T. Any one of those 20 can be picked for car number 1. Suppose for example you pick car K for the first one. Once you've picked that, there are 19 cars left to choose from for the second selection (K is no longer available). So suppose you pick car A. Now for the last car you 18 left to choose from (since A and K are both gone). So the total number of ways you can put together a 3-car train is 20*19*18 = 6840.

I think this is the answer they are looking for. Note that in this answer a train built as AKF is considered to be different than a train that is assembled as KAF. The earlier posters have responded assuming that AFK and KAF would be considered as duplicates - hence they divide this answer by 6 (because any 3 car train can be arrangd in 6 different ways). But I don't think that's the intent of the problem.

tstockero · Feb 4, 2009, 11:04 AM

Thank you! That is much easier to understand.:)