The correct option is (B); 2454 1st Case: 2 pair of identical letters can be arranged in \(^3C_2\)\(\times\)\(\frac{4!}{2!2!}\) 2nd Case: 2 identical and different letters can be arranged in \(^3C_1\)\(\times\)\(^7C_2\)\(\times\)\(\frac{4!}{2!}\) 3rd Case: All different letter can be arranged in \(^8P_4\) \(\therefore\) Total number of arrangements \(^3C_2\)\(\times\)\(\frac{4!}{2!2!}\)+\(^3C_1\)\(\times\)\(^7C_2\)\(\times\)\(\frac{4!}{2!}\)+\(^8P_4\)=2454