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Step-by-Step Solution
Step 1: Identify the Letters and Repetitions
The word “GTWENTY” consists of the letters G, T, W, E, N, T, and Y. Notice that “T” appears twice. This repetition must be taken into account when counting permutations.
Step 2: Arrange All Letters in Alphabetical Order
In alphabetical order, these letters are: E, G, N, T, T, W, Y. Our goal is to find the position of “GTWENTY” among all permutations of these letters, listed as in a dictionary (lexicographic order).
Step 3: Count Permutations Starting with Letters Less than G
Letters less than G in {E, G, N, T, T, W, Y} are just E.
If the first letter is E, the remaining 6 letters {G, N, T, T, W, Y} can be arranged in
\frac{6!}{2!} = \frac{720}{2} = 360 ways
(divided by 2! because of the two T’s).
Therefore, 360 permutations will appear before any word starting with G.
Step 4: Fix the First Letter as G, Move to the Second Letter
Now fixing G at the start, the remaining letters are {E, N, T, T, W, Y}. The next letter in “GTWENTY” is T. We must count permutations starting with “G” followed by each letter that is smaller than T.
Letters smaller than T are E and N (2 letters). If either E or N is chosen as the second letter, the remaining 5 letters (including one repeated T) can be arranged in
\frac{5!}{2!} = \frac{120}{2} = 60 ways.
Total permutations for these scenarios = 2 \times 60 = 120 .
Running total so far = 360 + 120 = 480.
Step 5: Fix "G T", Move to the Third Letter
With “G T” fixed, the remaining letters are {E, N, T, W, Y}. The third letter in “GTWENTY” is W. We count permutations starting with “G T” and having a third letter less than W.
Letters less than W in {E, N, T, W, Y} are E, N, and T (3 letters). For each of these as the third letter, the remaining 4 letters can be arranged in 4! = 24 ways.
Hence, total permutations before “G T W” = 3 \times 24 = 72 .
Running total = 480 + 72 = 552.
Step 6: Fix "G T W", Fourth Letter is E
Now “G T W” is fixed. The leftover letters are {E, N, T, Y}. The next letter in our target word is E. Among {E, N, T, Y}, no letter is smaller than E.
So, 0 permutations come before “G T W E”.
Running total remains 552.
Step 7: Fix "G T W E", Fifth Letter is N
The remaining letters are now {N, T, Y}. Alphabetically, these are N, T, Y. No letter among these is smaller than N.
So, 0 permutations come before “G T W E N”.
Running total stays 552.
Step 8: Fix "G T W E N", Sixth Letter is T
The leftover letters are {T, Y}. There is no letter smaller than T in {T, Y}.
So, 0 permutations come before “G T W E N T”.
Running total remains 552.
Step 9: The Final Letter is Y
Having accounted for all permutations that appear before “G T W E N T Y,” we add 1 for the position of “GTWENTY” itself:
Final Rank = 552 + 1 = 553.
Final Answer
The rank (dictionary position) of "GTWENTY" is 553.
