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