Number of Combinations

jason88 · #1 $Old$ August 21st, 2008, 12:04 AM

G'day,

not sure if this the right place to post this...

I have 20 dice/triangle-like objects, each with a 1, 2 and 3 on it.

How many possible combinations can be made from the 20 objects? such as 20x1 or 5x1,7x2, 8x3.

Any help greatly appreciated.
Cheers.

wisterville · #2 $Old$ August 21st, 2008, 01:13 AM

Hello,

You divide 20 (identical) dices into 3 categories: 1, 2 or 3.

So,
${}_3H_{20}={}_{3+20-1}C_{20}={}_{22}C_{20}={}_{22}C_2=\frac{22\cdot 21}{2!}=231.$

Maybe this notation is not so common, so let me explain a bit.
${}_{22}C_{20}$ is the binomial coefficient, sometimes written $\begin{pmatrix}22\\20\end{pmatrix}$ .
It counts the combinations, allowing no repetitions.
(You choose 20 different elements out of 22 ones. The order of choice is arbitrary.)
${}_{3}H_{20}$ counts the combinations, allowing repetitions.
(You choose 20 (maybe same)elements out of 3 ones. The order of choice is arbitrary.)

You align the 20 dices and (3-1=)2 separators in one row.
The dices left of the separators are 1,
the dices between the separators are 2,
the dices right of the separators are 3.
How many alignments are there?
You have 22 places to put the dices or separators,
choose 20 places for the dices and you are done.

Bye.

jason88 · #3 $Old$ August 21st, 2008, 01:23 AM

Quote:

Originally Posted by wisterville $View Post$

Hello,

You divide 20 (identical) dices into 3 categories: 1, 2 or 3.

So,
${}_3H_{20}={}_{3+20-1}C_{20} ={}_{22}C_{20}={}_{22}C_2 =\frac{22\cdot 21}{2!}=231.$

Maybe this notation is not so common, so let me explain a bit.
${}_{22}C_{20}$ is the binomial coefficient, sometimes written $\begin{pmatrix}22\\20\end{pmatrix}$ .
It counts the combinations, allowing no repetitions.
(You choose 20 different elements out of 22 ones. The order of choice is arbitrary.)
${}_{3}H_{20}$ counts the combinations, allowing repetitions.
(You choose 20 (maybe same)elements out of 3 ones. The order of choice is arbitrary.)

You align the 20 dices and (3-1=)2 separators in one row.
The dices left of the separators are 1,
the dices between the separators are 2,
the dices right of the separators are 3.
How many alignments are there?
You have 22 places to put the dices or separators,
choose 20 places for the dices and you are done.

Bye.

Thanks wisterville for your very informative answer.
I had no idea it was that complex! And there I was writing out all the combinations on paper!
how do you enter that forumla into a calculator?

Also, what if instead of dice:
there were 20 people who could each either run, skip or jump.
How many combinations can be made?

To be even more complicated, what if 10 people out of the 20 could only run and skip?

Thanks for any help.

wisterville · #4 $Old$ August 21st, 2008, 02:04 AM

Hello,

It is not so complicated once you get it...

For using the calculator, the formulae
${}_nH_r={}_{n+r-1}C_r$ and ${}_nC_r=\frac{n!}{(n-r)!r!}$ and $r!=r\cdot(r-1)\cdots 3\cdot2\cdot 1$ should do.

If the 20 dices were replaced by 20 people, you can distinguish between them so it becomes a permutation. In this case, you choose 20 (maybe same)elements out of 3 ones (The order of choice is not arbitrary).
Each person can choose from 3 alternatives, so ${}_3\Pi_{20}=3^{20}$ .

If 10 people have only 2 alternatives, the total would be
${}_2\Pi_{10}\cdot{}_3\Pi_{10}=2^{10}\cdot 3^{10}$ .

Bye.

jason88 · #5 $Old$ August 21st, 2008, 03:33 AM

thanks again wisterville.

wow, never would have thought there would be over 3 billion combinations (seems very high).

I would have been writing for a long time!