1. **State the problem:** We want to find the number of ways to arrange the letters of the word "AFRICA" such that the consonants are separated by at least one vowel. 2. **Identify letters:** The word "AFRICA" has 6 letters: A, F, R, I, C, A. - Vowels: A, I, A (3 vowels, with A repeated twice) - Consonants: F, R, C (3 consonants) 3. **Arrange vowels first:** Arrange the vowels A, I, A. - Number of ways to arrange vowels = \frac{3!}{2!} = 3 ways (since A repeats twice) 4. **Place consonants:** To ensure consonants are separated, place them in the slots between vowels. - For 3 vowels, there are 4 slots: _ V _ V _ V _ - We need to place 3 consonants in these 4 slots, one consonant per slot to keep them separated. - Number of ways to choose slots for consonants = \binom{4}{3} = 4 ways 5. **Arrange consonants:** The consonants F, R, C are distinct. - Number of ways to arrange consonants = 3! = 6 ways 6. **Calculate total arrangements:** - Total = (ways to arrange vowels) \times (ways to choose slots) \times (ways to arrange consonants) - Total = 3 \times 4 \times 6 = 72 **Final answer:** There are $72$ ways to arrange the letters of "AFRICA" such that the consonants are separated.