# Six boys and six girls are to sit in a row at random. The probability that boys and girls sit alternately is

$

A.{\text{ }}\dfrac{{6! \times {}^7{P_6}}}{{12!}} \\

B.{\text{ }}\dfrac{{6!6!}}{{12!}} \\

C.{\text{ }}\dfrac{{2 \times 6!6!}}{{12!}} \\

D.{\text{ }}\dfrac{{2 \times 6!}}{{12!}} \\

$

Last updated date: 22nd Mar 2023

•

Total views: 304.2k

•

Views today: 7.82k

Answer

Verified

304.2k+ views

Hint: First find out how many combinations of boys and girls sitting alternately are possible. Out of all the possible combinations girls would have 6! Ways because no. of girls are 6. But boys will have 7! (by using the concept of permutation) ways because the last boy can have two places either before the first girl or after the last girl and then accordingly calculate the probability.

Complete step-by-step answer:

Total ways in which all the 12 people sit : 12!

$n\left( s \right) = 12!$ ways where S is sample space.

E= Event of the girls and boys to be seated alternately.

6 girls can be seated in 6! ways.

${G_1},{G_2},{G_3},{G_{_4}},{G_5},{G_6}$

There are 7 alternate places for 6 boys.

6 boys can be seated in 7 places in ${}^7{P_6}$ ways.

$\therefore $ $n\left( E \right) = {\text{ 6! }} \times {}^7{{\text{P}}_6}$

$\therefore $ The required probability = $\dfrac{{n\left( E \right)}}{{n\left( S \right)}}$= \[\dfrac{{6!\; \times {}^7{P_6}}}{{12!}}\]

The correct option is A.

Note: The formula for calculating the permutation is given by \[P(n,r) = \dfrac{{n!}}{{(n - r)!}}\], where n is the total no. of object in the set, r is the no. of choosing object from the set.

Complete step-by-step answer:

Total ways in which all the 12 people sit : 12!

$n\left( s \right) = 12!$ ways where S is sample space.

E= Event of the girls and boys to be seated alternately.

6 girls can be seated in 6! ways.

${G_1},{G_2},{G_3},{G_{_4}},{G_5},{G_6}$

There are 7 alternate places for 6 boys.

6 boys can be seated in 7 places in ${}^7{P_6}$ ways.

$\therefore $ $n\left( E \right) = {\text{ 6! }} \times {}^7{{\text{P}}_6}$

$\therefore $ The required probability = $\dfrac{{n\left( E \right)}}{{n\left( S \right)}}$= \[\dfrac{{6!\; \times {}^7{P_6}}}{{12!}}\]

The correct option is A.

Note: The formula for calculating the permutation is given by \[P(n,r) = \dfrac{{n!}}{{(n - r)!}}\], where n is the total no. of object in the set, r is the no. of choosing object from the set.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE