Table Arrangement Problem with Combinations

Hello,

I am trying to help someone with this problem. I have absolutely no clue where to even begin in this question. I am only asking for a nudge in the correct direction, not asking for anyone to solve the question completely.

"8 digits (0-7) are to be arranged in a 2 x 4 table. In how many ways can they be arranged if entry in row 1 must be less than row 2?"

Any and all help is greatly appreciated!

Thank you!

Re: Table Arrangement Problem with Combinations

Hey ianm.

Can you use a computer to check this for you? A simple algorithm should compute all possibilities for you in a matter of seconds.

Another way to do this is to set up a probability calculation where the limits are such that Row 1 < Row 2 but this is going to be a bit more involved.

Re: Table Arrangement Problem with Combinations

Quote:

Originally Posted by

**chiro** Hey ianm.

Can you use a computer to check this for you? A simple algorithm should compute all possibilities for you in a matter of seconds.

Another way to do this is to set up a probability calculation where the limits are such that Row 1 < Row 2 but this is going to be a bit more involved.

Hi,

I am helping someone in high school. The only available tools we have are a TI-83 calculator. Is the only option through using the computer? Personally I have R, though I am weak at it. I just assume there's another way, because on an exam, a grade 12 student will not be using R.

Cheers and thanks for the rapid reply!

Re: Table Arrangement Problem with Combinations

The other way is through exhaustion: you fix a particular number and then work through the possibilities.

A computer could do this for you and simply count the number of matches it gets to your criteria.

Either way, both simply evaluate the possibilities and then see if they match the criteria.

The thing that makes this complex is that you have dependencies between rows and columns (since you can't repeat elements).

Re: Table Arrangement Problem with Combinations

Hello,

He asked one of his peers, who has: $\displaystyle \frac{8!}{2^4} = 2520 $ for an answer, with the reasoning that there are 8! ways of arranging the 8 numbers, but you have divide by $\displaystyle 2^4 $, since this is the number of ways to permutate the rows where the first is less than the next column.

Would you agree with this?

For interest's sake, would you use R to calculate this?

Re: Table Arrangement Problem with Combinations

I'm not exactly sure on the specifics of the argument he gives so I can't comment on that solution.

I would indeed use R to calculate this and the number of iterations in the loop would be 8! which would be evaluated in a second or two.

The loop would simply evaluate a configuration, check if it satisfies the constraints and increment a counter if it meets them.

Re: Table Arrangement Problem with Combinations

Quote:

Originally Posted by

**chiro** I'm not exactly sure on the specifics of the argument he gives so I can't comment on that solution.

I would indeed use R to calculate this and the number of iterations in the loop would be 8! which would be evaluated in a second or two.

The loop would simply evaluate a configuration, check if it satisfies the constraints and increment a counter if it meets them.

Know of any really good ways to start learning R? Would this even be that hard in R?

Re: Table Arrangement Problem with Combinations

If you know something like C then R will be straight-forward.

You can just set variables equal to things without declaring them.

Just read up on loops and comparisons (for statement for loop and if statement for checking).

Re: Table Arrangement Problem with Combinations

Quote:

Originally Posted by

**ianm** Hello,

I am trying to help someone with this problem. I have absolutely no clue where to even begin in this question. I am only asking for a nudge in the correct direction, not asking for anyone to solve the question completely.

"8 digits (0-7) are to be arranged in a 2 x 4 table. In how many ways can they be arranged if entry in row 1 must be less than row 2?"

Any and all help is greatly appreciated!

Thank you!

There are $\displaystyle \frac{8!}{(2!)^4}$ ways to break the 8 digits into 4 subsets of size 2. If we arrange the digits in each pair so that the smaller is on top, these pairs form the columns in your table. For each such set of columns, the columns can be ordered in 4! ways. So in all, there are

$\displaystyle \frac{8!}{(2!)^4} \;4!$

possible arrangements.

Re: Table Arrangement Problem with Combinations

Oops, I replied too hastily on that one.

$\displaystyle \frac{8!}{(2!)^4}$ takes into account the ordering of the pairs as well as their selection, so there is no need to multiply by 4!. The number of acceptable tables is

$\displaystyle \frac{8!}{(2!)^4}$