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A person invites 8 guests to a dinner and places 5 of them at one table and the remaining 3 at another, both the tables being round. The number of ways in which the guests can be arranged is
Options:
Solution:
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Verified Answer
The correct answer is:
2688
The number of ways to select 5 person out of 8 is ${ }^8 C_5$. Now, number of ways to place 5 on round table is 4 ! and number of ways to place 3 on round table is 2 !.
So, required number of ways $={ }^8 C_5 \times 4 ! \times 2 !$
$$
=\frac{8 \times 7 \times 6}{3 \times 2} \times 4 \times 3 \times 2 \times 2=42 \times 64=2688
$$
Hence, option (b) is correct.
So, required number of ways $={ }^8 C_5 \times 4 ! \times 2 !$
$$
=\frac{8 \times 7 \times 6}{3 \times 2} \times 4 \times 3 \times 2 \times 2=42 \times 64=2688
$$
Hence, option (b) is correct.
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