Question
Question: 5 boys and 3 girls sit in a row at random. The probability that no girls sit together is A. \(\dfr...
5 boys and 3 girls sit in a row at random. The probability that no girls sit together is
A. 143
B. 145
C. 149
D. 147
Solution
We will first find the total number of ways in which all people can be arranged. Then, find the number of ways in which 3 girls can sit together by considering all three girls as a single unit. And at last, subtract the number of ways in which three girls together from the total number of ways in all persons can be arranged.
Complete step-by-step answer:
When there are n objects and n places, then the number of ways in which objects can be arranged is n!.
There are total 8 person and 8 seats, then the total number of ways in which the person can be seated is 8!
Also, we know n!=n.(n−1).(n−2).....3.2.1.
Then, total possible outcomes are 8!=8.7.6.5.4.3.2.1=40320,
But, we need that no two girls sit side by side.
Then, there has to be a girl between every two boys.
Here, we can see there are 6 positions that a girl can occupy.
Then, 3 girls can choose any of the six positions.
We will use combinations to find the number of ways in which 3 girls can be seated when no two girls can sit together.
Hence, there are 6C3
Also, nCr=r!(n−r)!n!
Then, 6C3=3!(3)!6!=3.2.1.3!6.5.4.3!=20
Also, the three girls and 5 boys can arrange among themselves.
Therefore, total ways in which two girls can sit such that no girls sit together.
5!3!(20)=5.4.3.2.1.3.2.1.20=14400
We will calculate the probability by dividing the number of favourable outcomes by the total number of all the possible outcomes.
4032014400=145
Hence, option B is correct.
Note: When there are n distinct objects and needs to be arranged in n places, then the number of ways in which these can be arranged is n!. Also, n!=n.(n−1).(n−2).....3.2.1. Since, the position of the girls and boys will also matter that’s why we have to multiply 3! and 5! by finding the total ways of arrangement.