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 stand in a row so that no two men of same nationality are next to one another

• A. $47\times 8!$
• B. $10!$
• C. $2!.2!.2!.4!$
• D. none of these

Answer 1

Answer: (A)

$47\times 8!$

Suppose $2$ objects are to be kept apart in a permutation, there are $2$ methods

The "gap method", where these $2$ objects are placed in the gaps (incl. ends)

The "subtraction method" [ total permutations - those with the $2$ objects together ]

By successively applying these, we can find the number of ways both A′s and B′s are apart, and then place the C′s, as separators, if so needed.

A′s apart or together, B′s apart $=6!\times 7\cdot 6=6!\times 42$

(a) A′s together, B′s apart $=(2\times 5!)\times 6\cdot 5=6!\times 10$

(b) thus both A′s and B′s apart $=6!(42-10)=6!\times 32$

(c) also, both A′s and B′s together $=6!\times 4$

Now placing the C′s appropriately, remembering that the two are distinct, and that there will be two versions of case (a), we get

$6!\left[\right(2\times 10)\times (2\cdot 8)+(32\times 8\cdot 9)+(2\cdot 4\left)\right]=8!\times 47$