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#131 Re: Help Me ! » 15 people in a row » 2016-04-19 13:43:15

For completeness;





The general formulas are somewhat elusive this time. I have only got:

For permutations with 2+ adjacent females:

A generating function for permutations with 3+ adjacent males:

With recursive formula:

And the previous wrong formulas

#132 Re: Help Me ! » 15 people in a row » 2016-04-14 17:54:25

The values are correct. Once I switched to Excel everything matched (up to 12 seats so far). I will do the rest later and attempt general rules. Also keeping track of the numbers of adjacent and multiple adjacent combinations.

Good to finally have a reliable count.

#133 Re: Help Me ! » How do you solve an equation like this? » 2016-04-14 01:10:43

It would be interesting to be an expert in the representation of solutions with no analytic expression. Surely there is in principle some expression with no self-reference, if everything including limits, integrals, functions etc. are included?

#134 Re: Help Me ! » How do you solve an equation like this? » 2016-04-13 22:57:55

Hi bobbym,

The only difference between the forms in post #2 and #1 is that #2 is continuous at x=0.

Thank you for bringing integer relation algorithms to my attention! big_smile

It has honestly never occurred to me before that there might be numbers that can't be expressed in terms of known constants. It sounded counter-intuitive to me, until I realised that it does not imply that there are numbers which cannot be expressed in terms of defined operations. It merely means that it is sometimes necessary to expand our collection of "known constants".

#136 Re: Help Me ! » How do you solve an equation like this? » 2016-04-13 21:24:33

It does not have to be done by hand, but I would like to know something about the two analytical forms (or determining that they cannot be found). If you cannot find them, can you elaborate on how this might be investigated?

It is not a practical problem since it is relatively easy to approximate the answers to a desired degree. I am just curious (and pedantic) tongue

#138 Help Me ! » How do you solve an equation like this? » 2016-04-13 15:47:10

Relentless
Replies: 8

I don't have the time to figure out how to do it right now, so I thought I might post it while I'm here.
It gives the solutions for the discount rates per annum that set two engineering options equal in terms of present financial value. It is not an assessment question, it is just something I encountered. Exact solutions are sought.

It is already simplified a little, but I am a bit lost now.

I am concerned more with answers than explanations at the moment. (:

#139 Re: Help Me ! » 15 people in a row » 2016-04-13 13:51:18

Where does the huge disparity come from? What is missing? :S

If a whole lot more is missed in the adjacent combinations, then this calculator is getting really annoying xD

Please share what numbers you have so that I have a chance to derive general rules and figure out what is going on.

#144 Re: Help Me ! » 15 people in a row » 2016-04-12 17:34:57

From all of the new formulas, the updated general formula for the answer in terms of n seats is:

The updated answer for n=15:


(about 51.19% of the total arrangements)

And we are as far as ever from an intuitive explanation smile



#145 Re: Help Me ! » 15 people in a row » 2016-04-12 14:52:16

Fortunately, the patterns became evident almost immediately.

I believe the real formula for number of combinations with 2 adjacent women for n number of seats is:

For three adjacent males:

In general, for the number of combinations with x adjacent types for n seats:

I will post the new duplicates between two females and three males later, and then we are basically done.





The patterns for correcting the number of duplicates also become evident after some calculation. It appears the correct number of duplicates is

in terms of the previous formula
This time I cross-referenced all combinations of *,m,m,m,* and m,*,m,m,m with *,w,w,* and w,*,w,w and manually (with a spreadsheet) counted the unique combinations.
To derive a formula for the numbers thus obtained from the formula we have, I believe the correct expression for the number of combinations with both two adjacent females and three adjacent males, for n seats, is:

(this is not applicable to one or two seats)

#146 Re: Help Me ! » 15 people in a row » 2016-04-12 13:15:27

Oh my gosh, you are correct. 21 is definitely right. But I haven't the faintest why the darn thing isn't picking those up.

*,w,w,* fails to search w,*,w,w. How... irritating smile

Similarly 5 seats is missing
{w,k,k,w,w}
{w,k,m,w,w}
{w,k,w,w,k}
{w,k,w,w,m}
{w,k,w,w,w}
{w,m,k,w,w}
{w,m,m,w,w}
{w,m,w,w,k}
{w,m,w,w,m}
{w,m,w,w,w}

so the right number is 75 instead of 65...

The problem is not too difficult to overcome, however. I will list the numbers of combinations of w,*,w,w as well as the numbers of duplicates between w,*,w,w and *,w,w,* shortly. All we have to do is add the first and subtract the second from our given numbers and general rule. Since the result then agrees so far with manual and exhaustive listing, this should be the only quirk.

#147 Re: Jai Ganesh's Puzzles » Series and Progressions » 2016-04-12 02:52:42

Hello big_smile

You can find how many integers x there are divisible by y within a range by choosing a high number a and a low number b, at least one of which is divisible by y, solving this equation, and flooring (rounding down) the result:

Equivalently,

a and b are excluded from the count

Finally, IF BOTH a AND b are divisible by y, subtract 1.

Surely there are also more elegant methods xD

#150 Re: Help Me ! » 15 people in a row » 2016-04-12 02:15:15

Hi (:

The list does include {w,w,w,m} and {w,w,w,w} in the 9th and 10th rows.
I used "pattern *,w,w,*" for this.
"pattern *,m,m,m,*" for the males,
and both for both.

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