The word NET has three distinct letters: N, E and T. The number of permutations of n distinct objects taken all at a time is n!, so here it is 3!. Calculating 3! gives 3 × 2 × 1 = 6. Therefore, there are 6 possible distinct arrangements of the letters in NET.
Option A:
Option A uses the factorial formula for permutations correctly, recognizing that there are no repeated letters. By computing 3! as 6, it counts all possible orderings such as NET, NTE, ENT, ETN, TEN and TNE.
Option B:
Option B, 3, would be the number of letters, not the number of arrangements. It undercounts the permutations by ignoring the different orders in which letters can appear.
Option C:
Option C, 9, may come from incorrectly squaring the number of letters (3²), which does not match the combinatorial rule for arrangements of distinct objects.
Option D:
Option D, 12, overestimates the possible arrangements and could arise from misapplying permutation formulas that involve repetition or additional positions.
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