The number of ways in which 10 candidates A 1 , A 2 , . . ., A 10 can be ranked so that A…

Mathematics · JEE Advanced · NTA ExamsPermutations and Combinations

The number of ways in which 10 candidates A1, A2, . . ., A10 can be ranked so that A1 is always before A2 is :
  1. \(\frac{0!}{2}\)
  2. 8! × 10C2
  3. 10P2
  4. 10C2

Answer

(A) 0! 2, (B) 8! × 10 C 2

Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.

Related practice questions

More Permutations and Combinations questions · Browse all practice questions