how many different combinations are there if i have 6 boxes with 4 different items and take 1 from each?

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- Jan 23rd 2009, 07:30 AMjplondonpossible combinations??
how many different combinations are there if i have 6 boxes with 4 different items and take 1 from each?

- Jan 23rd 2009, 07:53 AMPlato
The way you have stated the question, it has many interpretations.

Please clarify the question.

Are the boxes different in content?

Does each box have exactly the same number of each of the four items?

For example: if the items are A, B, C, & D, does each box have exactly those four letters?

Can you please clear away the confusion. - Jan 23rd 2009, 09:42 AMjplondon
Thank you for your response, I will try to explain more clearly

I work for a printing company, we have a machine the feeds individual cards onto a conveyor and wraps them in plastic.

In this case we have 24 different cards. we will shuffle and load cards 1 - 4 in feeder 1, 5 - 8 in feeder 2 etc. so we end up with 6 feeders containing 4 different cards in each, 24 different cards in total.

I want to tell the customer how many different pack versions there will be.

Hope this is a bit clearer - Jan 23rd 2009, 11:03 AMPlato
It is a lot clearer. But I know it must be me, I am still unsure.

So let tell you this much. If we were to divide 24 individuals into six groups of four each that can be done in $\displaystyle \frac{{24!}}{{\left( {4!} \right)^6 \left( {6!} \right)}} = {\rm{4,509,264,634,875}}$ ways.

That is how many different packets of four different cards, 1 to 24, are possible.

The bits about a conveyor and a feeder still confuse.

If this is still not it, I will be glad to give it another try. - Jan 23rd 2009, 11:10 AMSoroban
Hello, jplondon!

If I read the problem correctly, it is quite complicated.

Quote:

How many different combinations are there if i have 6 boxes

with 4 different items in each box, and take 1 from each box

If you**don't**shuffle the 24 cards before loading the six feeders,

. . the problem is much simpler.

The arrangement might look like this:

. . $\displaystyle \begin{array}{ccc}\boxed{1,2,3,4} & \boxed{5,6,7,8} & \boxed{9,10,11,12} \\ \\ \boxed{13,14,15,16} & \boxed{17,18,19,20} & \boxed{21,22,23,24} \end{array}$

We take one from each feeder and for each there are 4 choices.

Hence, there are: .$\displaystyle 4^6 \:=\:4096$ possible four-card packs.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you**do**shuffle the 24 cards before leading the feeders,

. . there will be far more possible outcomes.

But there will so much duplication, that it is virtually impossible

. . to come up with an accurate count.

In the first arrangement, we could select: .$\displaystyle \{4,8,12,16,20,24\}$

In one of the shufflings, the distribution might be, for example:

. . $\displaystyle \begin{array}{ccc}\boxed{1,{\color{red}8},10,15} & \boxed{5,{\color{red}12},17,22} & \boxed{{\color{red}4},11,19,23} \\ \\ \boxed{2,6,14,{\color{red}20}} & \boxed{7,13,18,{\color{red}24}} & \boxed{3,9,{\color{red}16},21} \end{array}$

And we could still select $\displaystyle \{4,8,12,16,20,24\}$

I hope you see the difficulty . . .

- Jan 24th 2009, 12:52 PMawkward
Hi jplondon,

Let me state my understanding of the problem: You have 6 feeders, each of which contains 4 cards, so there are 24 cards in all (the cards in a feeder are all different from those in other feeders). You select one card from each feeder, and you would like to know in how many ways this can be done.

There are 4 ways to select the card from the first feeder, 4 ways to select the card from the second feeder, etc., so the answer is

$\displaystyle 4^6 = 4096$. - Jan 25th 2009, 12:59 AMGrandadCards problem
Hello jplondon -

I'm afraid I'm still not clear how many cards each customer gets.

Does he get a pack of 6, chosen at random from the 24 different ones? If he does, there are $\displaystyle ^{24}C_6$ different packs $\displaystyle = 134,596$ different packs.

Or does he get all 24 cards, arranged into 6 packs of 4 cards each? If this is it, Plato's answer is the one you want.

Or does he get all 24 cards, arranged into 4 packs of 6 cards each? If so, then there are $\displaystyle \frac{24!}{(6!)^4(4!)} =96,197,645,544$ possible combinations.

Take your pick!

Grandad