A man invites 6 non-vegeterian and 5 vegeterian friends for a dinner party. He arrange 6 non-vegeterian friends on one round table and 5 vegeterian friends along another round table. The number of ways this can be done is:

# A man invites 6 non-vegeterian and 5 vegeterian friends for a dinner party. He arrange 6 non-vegeterian friends on one round table and 5 vegeterian friends along another round table. The number of ways this can be done is:

1. A

$11!$

2. B

9!

3. C

2880

4. D

8280

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### Solution:

Arrange non-vegeterians in (6 -1)! =5! ways and vegeterian in (5 – 1)! = 4! ways

$\therefore$ required number of ways $=\left(5!\right)\left(4!\right)=\left(120\right)\left(24\right)=2880$

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