(D) Total number of experts $= 2 + 3 + 4 = 9$. Number of ways to select $3$ experts out of $9$ is ${}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \ti

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(D) Total number of experts $= 2 + 3 + 4 = 9$. Number of ways to select $3$ experts out of $9$ is ${}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$. We want to select $3$ experts such that they belong to different institutions,meaning one from $A$,one from $B$,and one from $C$. Number of ways to select one expert from each institution $= {}^2C_1 \times {}^3C_1 \times {}^4C_1 = 2 \times 3 \times 4 = 24$. Required probability $= \frac{24}{84} = \frac{2}{7}$.

Class 11 Mathematics Probability Advanced Use of permutations and combinations in probability

Medium

$A$ committee consists of $9$ experts taken from three institutions $A, B$,and $C$,of which $2$ are from $A$,$3$ from $B$,and $4$ from $C$. If three experts resign,then the probability that they belong to different institutions is

A
$\frac{1}{729}$
B
$\frac{1}{24}$
C
$\frac{1}{21}$
D
$\frac{2}{7}$

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Probability

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Advanced Use of permutations and combinations in probability

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$A$ committee consists of $9$ experts taken from three institutions $A, B$,and $C$,of which $2$ are from $A$,$3$ from $B$,and $4$ from $C$. If three experts resign,then the probability that they belong to different institutions is

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