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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Permutation and Combination

BITSAT 2020 · Mathematics · Permutation and Combination

A person invites a party of 10 friends at dinner and place them so...

A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is

Options

  1. A(10)!6!\frac{(10) !}{6 !}
  2. B(10)!24\frac{(10) !}{24}
  3. C(9)!24\frac{(9) !}{24}
  4. DNone of these

Correct answer

B. (10)!24\frac{(10) !}{24}

Step-by-step solution

Selection of 6 guests =10C6=10 \mathrm{C} 6 Permutation of 6 on round table =5!=5 ! Permutation of 4 on round table =3!=3 ! Then, total number of arrangements =10C6.5!.3!=(10)!6!4!⋅5!.3!=(10)!24=10 \mathrm{C} 6.5 !.3 !=\frac{(10) !}{6 ! 4 !} \cdot 5 !.3 !=\frac{(10) !}{24}
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