Question

In a party there are $2$ Americans, $2$ Britishmen, $2$ chinese and four men of other nationalities. Find the number of ways in which they can sit around a table such that no two men of same nationality are next to one another

• A. $244\times 6!$
• B. $10!$
• C. $2!.2!.2!.4!$
• D. none of these

Answer 1

The correct option is A $244\times 6!$

Total no. of ways they can be seated

$=9!=504(6!)$

$2$ Americans, $2$ Chinese, $2$ British, $4$ others

Let, $A-2$ Americans together

$B-2$ British together

$C-2$ Chinese together

$\therefore$ No of ways in which atleast 2people of same nationality are together is $n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)$

$=3[8!2!]-3[7!2!2!]+6!2!2!2!=6!\left[260\right]$

$\therefore$ Req no. of ways $=6![504-260]=244(6!)$