SATHEE: Permutations And Combinations Question 336

Permutations And Combinations Question 336

Question: 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:

A) $ \frac{(10)!}{6!} $

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

C) $ \frac{(9)!}{24} $

D) None of these

Show Answer

Answer:

Correct Answer: B

Solution:

[b] Selection of 6 guests = 10C6 Permutation of 6 on round table = 5! Permutation of 4 on round table = 3! Then, total number of arrangements $ {{=}^{10}}C_5.5!.3! $ $ =\frac{(10)!}{6!4!}.5!.3!=\frac{(10)!}{24} $

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Permutations And Combinations Question 336

Question: 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:

Answer:

Solution:

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