A party of $10$ consists of $2$ Americans, $2$ Britishmen, $2$ Chinese and $4$ men of other nationalities(all different). Let the number of ways in which they can stand in a row so that no two men of the same nationality are next to one another be $(k) m!$ and also the number of ways in which they can sit at a round table be $(p) q!$ . Find $\frac{p + q}{k + m - 5}$ ?

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Solution

(i) Total ways $= 10!$
Undesirable cases : When $2$ Americans are together $(A_{1} A_{2})$
or two British are together $(B_{1} B_{2})$ or two Chinese are together $(C_{1} C_{2})$
We plot them on Venn diagram :
we use,
$n (A_{1} A_{2} \cup B_{1} B_{2} \cup C_{1} C_{2}) = n (A_{1} A_{2}) + n (B_{1} B_{2}) + n (C_{1} C_{2}) - n [(A_{1} A_{2}) \cup (B_{1} B_{2})] - n [(B_{1} B_{2}) \cup (C_{1} C_{2})] - n [(C_{1} C_{2}) \cup (A_{1} A_{2})] + n [(A_{1} A_{2}) \cap (B_{1} B_{2}) \cap (C_{1} C_{2})]$

Where $n (A_{1} A_{2})$ denotes $\to$ when $2$ Americans are together $= 9! 2!$

Correspondingly for $B_{1} B_{2} & C_{1} C 2$
$n [(A_{1} A_{2}) \cup (B_{1} B_{2})]$ denotes when $2$ Americans and $2$ Britishmen are together $= 8! \times 2! \times 2!$
Correspondingly same for others
$n [(A_{1} A_{2}) \cap (B_{1} B_{2}) \cap (C_{1} C_{2})]$ denotes when $2$ Americans, $2$ Britishmen and $2$ Chinese are together $= 7! \times 2! \times 2! \times 2! = 8!$
Put values we get
$n (A_{1} A_{2} \cup B_{1} B_{2} \cup C_{1} C_{2})$
$= 9! \times 2! \times 3 - 8! \times 2 \times 2 \times 3 + 8!$
$= 8! (43)$
These are undesired ways
Desired ways $= 10! - 8! (43) = 8! (47) = (k) m!$
$k = 47, m = 8$
(ii) Now they are on a round table
Total ways $= (n - 1)! = (10 - 1)! = 9!$
Undesired ways :
$n (A_{1} A_{2} \cup B_{1} B_{2} \cup C_{1} C_{2})$
$= 8! \times 2 \times 3 - 7! \times 2! \times 2! \times 3 + 6! \times 2! \times 2! \times 2!$
$= 6! \times 4 [7 \times 2 \times 2 \times 3 - 7 \times 3 + 2]$
$= 6! \times 260$
Desired ways $= 9! - 6! \times 260 = (244) 6! = (p) q!$ ways
$p = 244, q = 6$

Therefore,
$\frac{p + q}{k + m - 5} = \frac{244 + 6}{47 + 8 - 5} = 5$

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