Reviewing, 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.)

Whyso? 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.

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  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, ...".)

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  In the first "Transmuting"file, I PERMUTED only LETTERS. I shall now PERMUTE COLORS. Since it is easier in the present format to INDICATE COLORS via LETTERS, I'll simply run through this previous file, with its LETTER GNOMONS as COLORED LETTERS.

  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.)

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  Next, we EXTEND and RECUR:

  1. we start with S₁ to GENERATE S₂: 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. I 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: ABC.
  3. We now "swing" C ACROSS this first ORDERING: ACB; and again, CAB.
  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 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, 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 it over: ABDC; SWING again, ADBC; SWING again, DABC.
  4. We've now brought the SWINGER 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: