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Standard XII

Mathematics

Circular Permutation

A man invites...

Question

A man invites 10 friends to a party and places 5 at one table and 5 at another table, the tables being round. The number of ways in which he can arrange the friends is

(4!)^{2}

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^{10} C_{5} (4!)^{2}

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^{10} C_{5} (5!)^{2}

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Solution

The correct option is B $^{10} C_{5} (4!)^{2}$
Number of ways selecting 5 out of 10 is $^{10} C_{5}$ and arranging them in circular table then number of arrangement of 5 persons in circualr table is $(5 - 1)!$
$∴$ required number of ways $=^{10} C_{5} (4!)^{2}$

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A man invites 10 friends to a party and places 5 at one table and 5 at another table, the tables being round. The number of ways in which he can arrange the friends is

The correct option is B 10C5(4!)2Number of ways selecting 5 out of 10 is 10C5 and arranging them in circular table then number of arrangement of 5 persons in circualr table is (5−1)!∴ required number of ways =10C5(4!)2

The correct option is B $^{10} C_{5} (4!)^{2}$
Number of ways selecting 5 out of 10 is $^{10} C_{5}$ and arranging them in circular table then number of arrangement of 5 persons in circualr table is $(5 - 1)!$
$∴$ required number of ways $=^{10} C_{5} (4!)^{2}$