Question
State True or False for the following statement:
Eighteen guests are to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on other side of the table.
The number of ways in which the seating arrangements can be made is $\frac{11!}{5!6!}(9!)(9!) .$

Answer

True.Solution:
Let two sides of the table be $A$ ans $B$ each having $9$ seats. Let on side $A,$ four particular guests and on side $B,$ three particular seated. Now for side $A,$ five more guests can be selected from remaining eleven guests in $^{11}C_5$ ways. Also on each side nine guests can be arranged in 9! ways. So total number of ways of arrangement $=\ ^{11}\text{C}_5​​\times\ 9!\times9!​​=\frac{11!}{5!\ 6!}(9!)(9!)$

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