<p>The number of arrangements of the letters of the word <strong>BANANA</strong> in which the two N's do not appear adjacently is:</p>
Step-by-Step Solution
Key Concept: Use the complement principle: count total arrangements, subtract arrangements where N's are adjacent (treat NN as one unit).
The word BANANA contains 6 letters: one B, three A's, and two N's.
**Step 1: Determine the total number of distinct arrangements of these letters.**
To count the total arrangements, we consider the 6 positions for the letters. There are 3 identical 'A's. The two 'N's are treated as distinct for the purpose of counting permutations that satisfy the adjacency condition.
The total number of arrangements is:
$$ \frac{6!}{3!} = \frac{720}{6} = 120 $$
**Step 2: Determine the number of arrangements where the two N's appear adjacently.**
Treat the two N's as a single block (NN). Now we are arranging 5 items: B, A, A, A, and the (NN) block.
The number of arrangements of these 5 items, considering the three identical 'A's, is:
$$ \frac{5!}{3!} = \frac{120}{6} = 20 $$
Within the (NN) block, the two N's can be arranged in $2!$ ways (e.g., $N_1 N_2$ or $N_2 N_1$). Therefore, the total number of arrangements where the N's are adjacent is:
$$ 20 \times 2! = 20 \times 2 = 40 $$
**Step 3: Determine the number of arrangements where the two N's do not appear adjacently.**
This is found by subtracting the number of arrangements where the N's are adjacent from the total number of arrangements.
$$ 120 - 40 = 80 $$
Correct Answer: C