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Exercise 6.3 · Q19

Q.If the letters of the word SCHOOL are arranged as in dictionary, then find the rank of the word SCHOOL.

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Build the rank of SCHOOL by, at each letter position, counting all arrangements that would start with an alphabetically smaller unused letter, summing across positions, then adding 1.

[!FORMULA] Letters of SCHOOL sorted alphabetically: C,H,L,O,O,SC, H, L, O, O, S (OO repeats twice). At each position, for every unused letter smaller than the actual letter placed there, the number of words that would result is (remaining letters count)!repetition factorial of the remaining multiset\dfrac{(\text{remaining letters count})!}{\text{repetition factorial of the remaining multiset}}. Rank =1+∑(such counts across all positions)=1+\sum(\text{such counts across all positions}).

  1. Position 1 (target S): remaining pool {C,H,L,O,O,S}\{C,H,L,O,O,S\}, letters smaller than S are C,H,L,OC,H,L,O.
    • Remove CC: remaining H,L,O,O,S⇒5!2!=60H,L,O,O,S\Rightarrow \dfrac{5!}{2!}=60
    • Remove HH: remaining C,L,O,O,S⇒60C,L,O,O,S\Rightarrow 60
    • Remove LL: remaining C,H,O,O,S⇒60C,H,O,O,S\Rightarrow 60
    • Remove OO: remaining C,H,L,O,SC,H,L,O,S (only 1 OO left) ⇒5!=120\Rightarrow 5!=120 Subtotal =60+60+60+120=300=60+60+60+120=300.
  2. Fix position 1 =S=S. Remaining pool {C,H,L,O,O}\{C,H,L,O,O\}. Position 2 (target C): no letter in the pool is smaller than CC. Subtotal =0=0.
  3. Fix position 2 =C=C. Remaining pool {H,L,O,O}\{H,L,O,O\}. Position 3 (target H): no letter smaller than HH remains. Subtotal =0=0.
  4. Fix position 3 =H=H. Remaining pool {L,O,O}\{L,O,O\}. Position 4 (target O): letter smaller than OO is LL. …
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