The number of ways in which 10 candidates A 1 , A 2 , . . ., A 10 can be ranked so that A…
Mathematics · JEE Advanced · NTA Exams — Permutations and Combinations
The number of ways in which 10 candidates A1, A2, . . ., A10 can be ranked so that A1 is always before A2 is :
- \(\frac{0!}{2}\)
- 8! × 10C2
- 10P2
- 10C2
Answer
(A) 0! 2, (B) 8! × 10 C 2
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