## Combining Elements: How Many Variants?

In making a multiform story, whether on the Web or using a procedural environment like Character Maker, we can easily find ourselves overwhelmed by the possible directions the story could take.

It is important to approach these possibilities in an organized manner.

### Examples of multiple combinations:

• What three items do I take to the dance?
• Which two characters should go on the adventure?
• Which of these six events has taken place?
• Which of these "state flags" (e.g. hungry, angry, amorous, jealous, etc.) is on?

### How to approach a story world with multiple variants:

• Make a table of all the combinations (see below).
• Decide which combinations are definitely not allowed, and how you will prevent them from occuring.
• Decide which are the most desirable, dramatic, and/or "winning" combinations (sometimes the dramatic and narratively interesting ones are the "losing" combinations).
• Determine what are the most important polarities, contrasts, juxtapositions. What information will be revealed one may but not another? What would cause an interactor to replay, retrace the story space?
• Decide whether or not there is a "default" situation: a condition in which the interactor will not have chosen actively, or will have taken the path of least resistance. Or perhaps you want to make one path that will unfold in the case of several combinations that are of lesser interest to you. Be sure the default combinations are clear to you and that the default path is clearly visible to the interactor.

### Making a Table of Possibilities

Even if you think you know all the combinations, it is helpful to methodically examine them. The interactor may expect combinations that you have overlooked. You may also want to reconsider which one you have included and which you have excluded.

Here are some sample tables that demonstrate the magnitude of possibilities that can arise from relatively few choices.

For choice of two elements out of five (assuming you can't choose the same element twice):

 A B C D E A AB AC AD AE B BC BD BE C CD CE D DE E

For choice of two out of n elements, result is (number of items-1) + (n-2) + (n-3)….[until n-n]

2 selections out of 5 elements yields 4+3+2+1=10 possibilities

But the generic formula is more complex. Where K= the number of items chosen, and N= the number of items in the complete set, the number of possible choices is calculated like this:

N!

___________

((N-K)!) (K!)

For the example above that would yield 5x4x3x2x1 =120 divided by

(3x2)(2) = 12

120/12 = 10

3 nonrepeating elements out of 5 would also yield 10 possible combinations:

 A B C D E AB ABC ABD ABE AC ACD ACE AD ADE AE BC BCD BCE BD BDE BE CD CDE CE DE

3 choices from 6 elements = 20 possible combinations

 A B C D E F AB ABC ABD ABE ABF AC ACD ACE ACF AD ADE ADF AE AEF BC BCD BCE BCF BD BDE BDF BE BEF CD CDE CDF CE CEF DE DEF

3 out of 8 elements yields 55 possibilities

 A B C D E F G H AB ABC ABD ABE ABF ABG ABH AC ACD ACE ACF ACG ACH AD ADE ADF ADG ADH AE AEF AEG AEH AF AFG AFH AG AGH AH BC BCD BCE BCF BCG BCH BD BDE BDF BDG BDH BE BEF BEG BEH BF BFG BFH BG BGH BH CD CDE CDF CDG CDH CE CDF CEG CEH CF CFG CFH CG CGH CH DE DEF DEG DEH DF DFG DFH DG DGH DH EF EFG EFH EG EGH EH FG FGH FH