1. **Problem Statement:** Find the number of words that can be formed with the letters of the word 'BANANARAMA' without changing the relative position of vowels and consonants. 2. **Identify vowels and consonants:** - Vowels in BANANARAMA: A, A, A, A, A (5 vowels) - Consonants in BANANARAMA: B, N, N, R, M (5 consonants) 3. **Relative position constraint:** The order of vowels and consonants must remain the same as in the original word. This means vowels must appear in the same sequence and consonants in the same sequence, but letters can be rearranged among themselves within vowels and consonants respectively. 4. **Count arrangements of vowels:** There are 5 vowels, all A's, so the number of distinct arrangements is $$\frac{5!}{5!} = 1$$ 5. **Count arrangements of consonants:** The consonants are B, N, N, R, M. Here, N repeats twice. Number of distinct arrangements is $$\frac{5!}{2!} = \frac{120}{2} = 60$$ 6. **Total number of words:** Since vowels and consonants are arranged independently but their relative positions are fixed, total number of words is $$1 \times 60 = 60$$ **Final answer:** 60