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14.Probability
medium
Let $n$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then the value of $\frac{m}{n}$ is
A
$5$
B
$6$
C
$7$
D
$8$
(IIT-2015)
Solution
$n =6!\cdot 5!$ ( 5 girls together arranged along with $5$ boys)
$m ={ }^5 C _4 \cdot(7!-2.6!) \cdot 4!$
($4$ out of 5 girls together arranged with others – number of cases all $5$ girls are together)
$\frac{ m }{ n }=\frac{5 \cdot 5 \cdot 6!\cdot 4!}{6!\cdot 5!}=5$
Standard 11
Mathematics
