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Step-by-Step Solution
Step 1: Arrange the Letters of "PUBLIC" in Alphabetical Order
The letters in "PUBLIC" are P, U, B, L, I, C. Alphabetically, they become:
B, C, I, L, P, U.
Step 2: Count All Permutations Starting with Letters that Come Before P
The first letter in "PUBLIC" is P. All letters that come before P in alphabetical
order are B, C, I, and L (4 letters). For each of these 4 possible first letters,
the remaining 5 letters can be arranged in
$5! = 120$ ways.
So, the total number of permutations with first letter B, C, I, or L is
$4 \times 5! = 4 \times 120 = 480$.
Step 3: Count All Permutations Starting with P and Second Letter that Comes Before U
Now fix P as the first letter. The remaining letters are U, B, L, I, C. In alphabetical
order, they are B, C, I, L, U. The letters that come before U here are B, C, I, and L (4 letters).
For each choice of second letter among these 4, the remaining 4 letters can be arranged
in $4! = 24$ ways.
Hence, the total permutations for this step are
$4 \times 4! = 4 \times 24 = 96$.
Cumulative total so far: $480 + 96 = 576$.
Step 4: Count All Permutations Starting with "PU" and Third Letter Before B
Next, fix the first two letters as "PU". The remaining letters are B, L, I, C. Alphabetically,
those are B, C, I, L. Since the third letter in "PUBLIC" is B, we look for letters before B in
the list (none). So there are no permutations to add here.
Cumulative total remains $576$.
Step 5: Count All Permutations Starting with "PUB" and Fourth Letter Before L
Now fix the first three letters as "PUB". The remaining letters are L, I, C. In alphabetical order,
they are C, I, L. For the fourth letter as L, the letters before L are C and I (2 letters).
For each of these 2 choices, the remaining 2 letters can be arranged in
$2! = 2$ ways.
So, the total permutations for this step are
$2 \times 2! = 2 \times 2 = 4$.
Cumulative total now: $576 + 4 = 580$.
Step 6: Count All Permutations Starting with "PUBL" and Fifth Letter Before I
Fix "PUBL" as the first four letters. The remaining letters are I, C. Alphabetically,
they are C, I. For the fifth letter in "PUBLIC" to be I, the letter before it is C (1 letter).
That choice (C as fifth letter) leaves exactly $1! = 1$ arrangement for the last letter.
So, the total permutations for this step are
$1 \times 1! = 1$.
Cumulative total now: $580 + 1 = 581$.
Step 7: Position of the Word "PUBLIC"
After these 581 permutations, the very next permutation is "PUBLIC" itself.
Therefore, its position is $581 + 1 = 582$.
Final Answer
The dictionary rank of "PUBLIC" among its permutations is 582.
