There are 11 players $x_{1}, x_{2}, x_{3}, ........... x_{11}$ in cricket team. Players $x_{1}, x_{2}, .............. x_{7}$ will bat at first seven places and players $x_{8}, x_{9}, x_{10}, x_{11}$ will bat at last 4 places. Let the number of ways a batting order of 11 players can be made such that $x_{3}$ does not play at number three, $x_{6}$ does not play at number 7 and $x_{9}$ does not play at number 9 be ‘K’. Then the value of $\frac{K}{10}$ is

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Detailed Solution

$\overset{L e f t}{x_{1} x_{2} x_{3} x_{4} x_{5} x_{6} x_{7} |} \overset{R i g h t}{x_{8} x_{9} x_{10} x_{11}}$

No. of ways in which batsman can be arranged if $x_{3}$ does not play at third place and $x_{6}$ does not play at number 7.
$= 7! - (6! + 6! - 5!) = 7 \times 6 \times 5! - 11 \times 5! = 31 \times 5!$
In right side of if $x_{9}$ does not play at no 9 then no of ways of arrangement
$= 4! - 3! = 18$
Total no of arrangement