Multiple choice

Number of permutations of all letters of word $"LEADER"$ so that no two same letters are together

  1. $\dfrac{6!}{2!}$
  2. $180$
  3. $240$
  4. $360$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

LEADER has six letters, with E repeated twice. The total number of distinct arrangements is 6!/2! = 360, and arrangements with adjacent Es number 5! = 120, so the required count is 360 - 120 = 240.

AI explanation

The word LEADER has 6 total letters, where E appears twice. The total number of permutations of all the letters is 6! divided by 2!, which equals 720 divided by 2 or 360. To find the permutations where the two E letters are never together, we subtract the arrangements where the E letters are treated as a single combined entity. The number of such together arrangements is 5 factorial, which is 120. The required number of permutations is therefore 360 minus 120, which equals 240.

Keep practicing — related questions