Question:

Step 1: Identify and classify the letters in "NARENDRABHAI" First, let's count the frequency of each letter in the word NARENDRABHAI: - N appears 2 times - A appears 3 times - R appears 2 times - E appears 1 time - D appears 1 time - B appears 1 time - H appears 1 time - I appears 1 time Total: 12 letters We classify these as: - **Repeated letters**: N(2), A(3), R(2) — totaling 7 letters - **Non-repeated letters**: E, D, B, H, I — totaling 5 letters Step 2: Arrange the repeated letters To ensure no two non-repeated letters are adjacent, we first arrange all the repeated letters. The number of ways to arrange 7 letters where N repeats 2 times, A repeats 3 times, and R repeats 2 times is: $$\frac{7!}{2! \cdot 3! \cdot 2!} = \frac{5040}{2 \cdot 6 \cdot 2} = \frac{5040}{24} = 210$$ Step 3: Identify gaps for placing non-repeated letters When we arrange 7 letters in a row, they create gaps where we can place the non-repeated letters to ensure no two non-repeated letters are adjacent. For 7 letters arranged in a line: _ L _ L _ L _ L _ L _ L _ L _ There are **8 gaps** (including the positions before the first letter and after the last letter). Step 4: Choose gaps and place non-repeated letters We need to select 5 gaps from these 8 available gaps to place our 5 non-repeated letters (one letter per gap). The number of ways to choose 5 gaps from 8 is: $$\binom{8}{5} = \frac{8!}{5! \cdot 3!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56$$ Step 5: Arrange the non-repeated letters in the chosen gaps The 5 non-repeated letters (E, D, B, H, I) can be arranged among themselves in: $$5! = 120 \text{ ways}$$ Step 6: Calculate the total number of arrangements The total number of valid arrangements is the product of: - Arrangements of repeated letters: 210 - Ways to choose gaps: 56 - Arrangements of non-repeated letters: $5!$ $$\text{Total} = 210 \times 56 \times 5! = 210 \times 56 \times 120$$ Step 7: Simplify to match the given options Let's express this in the form of the given options: $$210 \times 56 \times 5! = (210 \times 56) \times 5!$$ Notice that $210 = 42 \times 5$ and $56 = 56$, so: $$210 \times 56 = 11,760 = 98 \times 120 = 98 \times 5!$$ Therefore: $$\text{Total} = 98 \times 5! \times 5! = 98(5!)^2$$ **Final Answer: The number of words that can be formed is $\boxed{98(5!)^2}$, which corresponds to Option 4.**