  Question

$$18$$ guests have to be seated half on each side of a long table. $$4$$ particular guests desire to sit on one particular side and $$3$$ others on the other side. Determine the number of ways in which the sitting arrangements can be made.

A
(9!)2  B
11C4(9!)2  C
11C3(9!)2  D
11C5(9!)2  Solution

The correct option is C $$^{11}C_5 (9!)^2$$Let the two sides be $$A$$ and $$B$$. Assume that four particular guests wish to sit on side $$A$$. Four guests who wish to sit on side $$A$$ can be accommodated on nine chairs in $$_{ }^{ 9 }{ { P }_{ 4 } }$$ ways,and three guests who wish to sit on on side $$B$$ can be accommodated on nine chairs in $$_{ }^{ 9 }{ { P }_{ 3 } }$$ ways.Now, the remaining guests are left who can sit on $$11$$ chairs on both the sides of the table in $$(11!)$$ ways.Hence, the total number of ways in which $$18$$ persons can be seated,$$=_{ }^{ 9 }{ { P }_{ 4 } }\times_{ }^{ 9 }{ { P }_{ 3 } }\times(11!)$$$$=$$$$^{11}C_5 (9!)^2$$Mathematics

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