In how many ways can the letters of the word ELEVEN be arranged so that no two E's are next to each other?

First arrange the other letters L, V, N (all distinct): 3! = 6 ways.

These three letters create 4 gaps (including both ends). Place the three identical E's in three different gaps: C(4, 3) = 4 ways.

Total = 6 × 4 = 24.

Check by complement: Total arrangements = 6!/3! = 120 (E repeats 3 times); direct enumeration of the no-two-E-adjacent case confirms 24.