Q56 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics
11 players of the Indian cricket team are sitting at a circular table. The number of ways they can sit so that two players, the wicket keeper and the captain never sit together is ___
- A$10!$
- B$2 \times 9!$
- Correct answerC$8 \times 9!$
- D$9 \times 9!$
Solution
Total number of players is $11$.
Total number of ways to arrange $11$ players around a circular table is $(11 - 1)! = 10!$.
To find the number of ways where the wicket keeper and the captain sit together, consider them as a single unit.
The number of units to arrange is $9$ remaining players plus $1$ unit, which equals $10$ units.
The number of ways to arrange these $10$ units around a circular table is $(10 - 1)! = 9!$.
The wicket keeper and the captain can interchange their positions within the unit in $2! = 2$ ways.
The number of ways they sit together is $2 \times 9!$.
The number of ways they never sit together is the total number of ways minus the number of ways they sit together.
Required number of ways is $10! - 2 \times 9! = 10 \times 9! - 2 \times 9! = 8 \times 9!$.
Answer: $8 \times 9!$
Total number of ways to arrange $11$ players around a circular table is $(11 - 1)! = 10!$.
To find the number of ways where the wicket keeper and the captain sit together, consider them as a single unit.
The number of units to arrange is $9$ remaining players plus $1$ unit, which equals $10$ units.
The number of ways to arrange these $10$ units around a circular table is $(10 - 1)! = 9!$.
The wicket keeper and the captain can interchange their positions within the unit in $2! = 2$ ways.
The number of ways they sit together is $2 \times 9!$.
The number of ways they never sit together is the total number of ways minus the number of ways they sit together.
Required number of ways is $10! - 2 \times 9! = 10 \times 9! - 2 \times 9! = 8 \times 9!$.
Answer: $8 \times 9!$
Hence, the correct answer is option: 'C'.