Elsewhere, in "MATHEMATICAL "DNA", I cite (as essential math for Kids and Adults) Group Theory. GROUPS are behind (and within) all arts and crafts, behind (and within) every CONSERVATION LAW OF PHYSICS. Yet I can easily teach Kids THE GROUP CONCEPT in the creeping of a baby.

I can also teach Kids about the vast subject of FINITE GROUP THEORY ("infinite group theory" is another subject) via A SINGLE REPRESENTATIVE: THE SYMMETRIC GROUP.

Howso? Because EVERY FINITE GROUP (finite in members, millions in form) IS A SUBGROUP OF THE SYMMETRIC OR PERMUTATION CLASS OF GROUPS, hence, THE SYMMETRIC GROUP represents all finite groups.

We use "symmetry" to denote A PATTERN THAT REMAINS UNCHANGED WHEN WE CHANGE ITS PARTS OR ITS POSTION IN SPACE OR TIME. And PERMUTATION means INTERCHANGING MEMBERS IN AN ORDERING.

For example, under COLORED MULTIPLICATION PATTERNS, Kids learn that THE DIGITAL ROOT OF A NUMBER (SUM OF ITS DIGITS) IS INVARIANT (WITH UNCHANGING SYMMETRY) UNDER PERMUTATION OF THE DIGITS OF THE NUMBER.

Consider 261 = 9 x 29. Its other PERMUTATIONS ARE: 216 = 9 x 24; 126 = 9 x 14; 162 = 9 x 18; 612 = 9 x 68; 621 = 9 x 69.

Now, the DIGITAL ROOT OF ALL THESE NUMBERS IS: 2 + 6 + 1 = 2 + 1 + 6 = 1 + 2 + 6 = 1 + 6 + 2 = 6 + 1 + 2 = 6 + 2 + 1 = 9.

"Of course", you'll say, "addition is commutative and the sum is invariant in permuting the addends."

Yes. But you missed the point. WHEN THE DIGITAL ROOT OF DIFFERENT NUMBERS IS THE SAME, WE GET THE SAME REMAINDER WHEN WE DIVIDE BY 9. All these PERMUTATIONS ARE MULTIPLES OF 9, so would yield REMAINDER ZERO WHEN DIVIDED BY NINE! THAT PROPERTY DOESN'T CHANGE WHEN THE NUMBER IS WRITTEN IN DECIMAL NOTATION. BUT A NUMBER LOSES THAT PROPERTY WHEN IT IS WRITTEN IN SOME BASE OTHER THAN TEN -- SAY, BASE EIGHT! (This is shown elsewhere.)

I show Kids certain REPRESENTATIVE ORDERINGS such that:

• Each REPRESENTATIVE is a SYMMETRIC GROUP, denoted S_n, meaning that EACH ORDERING consists of n elements (letters, numbers, shapes, colors, tones, etc.).

• Given an INITIAL ORDERING, elements are interchanged or PERMUTED in ALL THE POSSIBLE WAYS. There's a COMBINATORIAL FORMULA FOR THE NUMBER OF ORDERINGS IN S_n.

• The number is "n!" ("n factorial"), the product: 1 x 2 x ... x n = n! ORDERINGS GENERATED FROM THE n ELEMENTS. (Thus, 1! = 1; 2! = 1 x 2 = 2; 3! = 1 x 2 x 3 = 6; 4! = 1 x 2 x 3 x 4 = 24; 5! = 1 x 2 x 3 x 4 x 5; etc. This last factorial means, for instance, that ONLY FIVE KIDS CAN FORM 120 DISTINCT BASKETBALL TEAMS! Since 9! = 362380, MERELY 9 KIDS CAN FORM 362380 BASEBALL OR SOFTBALL TEAMS! Tell them this the next time there's "nuthin tuh doo".)

Why a factorial of orderings involved? Because,

• given n distinct elements to choose from, YOU HAVE n CHOICES FOR 1ST CHOICE;

• THIS LEAVES n - 1 ELEMENTS FOR 2nd CHOICE.

• Since your CHOICES ARE INDEPENDENT (WHAT YOU CHOOSE DOESN'T DETERMINE YOUR NEXT CHOICE), you have n x (n - 1) ways of 1st & 2nd choices.

• You now have (n - 2) ways for a 3rd CHOICE, hence, n x (n - 1) x (n - 2) ways of choosing 3 elements in an ORDERING.

• Continuing in this fashion -- for ALL n CHOICES TO COMPOSE AN ORDERING -- you can COMPOSE AN ORDERING in n x (n - 1) x (n - 2) x ... x 3 x 2 x 1 = n! ways.

  ---

  I'll show how simple it is to obtain these PERMUTATIONS by

  1. showing how to use S₂ (symmetric group on 2 members), for RECURSIVE-GENERATION of S₃ (symmetric group on 3 members);

  2. then use S₃ to RECURSIVELY-GENERATE S₄.

  3. THEN, AS ONE OF THE TWO PARTS OF THE PRESENT PROBLEM, I'll ask you to use S₄ to ....

  This RECURSIVE-GENERATIVE METHOD is a simple example of the GNOMONING-RECURSING Strategy I'm pushing in MADMATH.

  1. The GNOMON of THE S_n GROUP is an ORDERING OF n DISTINCT ELMENTS.

  2. The GNOMON of THE ORDERING is ONE OF THE DISTINCT ELEMENTS. (Below, I'll use letters: "A, B, ...".)

  ---

  Let's begin with the initiator, S₁ (symmetric group on 1 member): A.

  That's it. We can't permute it any more, which agrees with our formula for 1! = 1 as the "size" of S₁.)

  ---

  Next, we EXTEND and RECUR:

  1. we start with S₁ to GENERATE S₂, and S₁ is A.

  2. To that first (and only) form of S₁, we ADJOIN A "SWINGER", B, for the first form of S₂, namely, AB.

  3. But AB can be PERMUTED, namely, BA.

  4. Is this all? Yes. Obviously. And, by our FORMULA, S₂ has 2! = 1 X 2 = 2 ORDERINGS:

  
   	                 AB
                           BA.

  ---

  We next RECURSIVE-GENERATE S₃ (symmetric group on 3 members) from S₂:

  1. Start with the first ORDERING of S₂: AB.

  2. To this, we ADJOIN a "swinger", C, to complete the first ORDERING of S₃: AB -> ABC.

  3. We now "swing" C LEFT-ACROSS the first ORDERING, obtaing: ACBCAB.

  4. We have now taken the "swinger" (C) from RIGHT to LEFT, COMPLETING THE SWINGS FOR THIS ORDERING FROM S₂.

  5. We now bring down the next ORDERING from S₂, namely, BA, to REPLACE THAT S₂ form we just found -- obtaining CBA.

  6. And we SWING C across this ORDERING: BCA; and again, BAC.

  7. We've now swung from LEFT to RIGHT; and we've exhausted the ORDERINGS from S₂ -- so we've completed S₃, satisfying our formula, 3! = 1 X 2 X 3 = 6 ORDERINGS for S₃.

  
  	      ABC   CBA
  	
  	      ACB   BCA
  
  	      CAB   BAC.

  ---

  Now, let's use those SIX MEMBERS OF S₃ TO GENERATE THE 24 MEMBERS OF S₄.

  1. We bring down the first form of S₃, namely, ABC.

  2. To this we adjoin a SWINGER, D to COMPOSE THE 1ST ORDERING OF S₄: ABCD.

  3. To obtain the 2nd ORDERING of S₄, we SWING D rightward: ABDC; SWING again, ADBC; SWING again, DABC.

  4. We've now brought the SWINGER (D) from RIGHT TO LEFT, so we bring down the 2nd form of S₃, and SWING through it.

  5. We continue with these steps until we've brought down the 6th form of S₃, and SWUNG through it -- OBTAINING THE 4! = 24 ORDERINGS of S₄:

  
              ABCD   DACB   CABD   DCBA   BCAD   DBAC
  
              ABDC   ADCB   CADB   CDBA   BCDA   BDAC
  
              ADBC   ACDB   CDAB   CBDA   BDCA   BADC
  
              DABC   ACBD   DCAB   CBAD   DBCA   bACD 

  ---

  ETSETTERY! DIG THEE TRANSMUTING?