Question

If there are 6 girls and 5 boys who sit in a row, then the probability that no two boys sit together is:
(a) \[\dfrac{6!6!}{2!11!}\]
(b) \[\dfrac{7!5!}{2!11!}\]
(c) \[\dfrac{6!7!}{2!11!}\]
(d) None of these

Answer 1

Hint: At first make all the girls sit down which can be done in \[{}^{6}{{P}_{6}}\] ways. Then we will make 1 boy sit between two girls which are available for 5 spaces and also two which are on extreme ends. Hence this can be done in \[{}^{7}{{C}_{5}}\times 5!\].

Complete step-by-step solution:
In the question we are given that there are six girls and five boys sitting in the row, we have to find the probability that they do not sit together.
So we were at first told about 6 girls so we will place them first in any manner which can be done in \[{}^{6}{{P}_{6}}\] ways.
Now after making the girls sit we have to sit five boys too such that none of them sit together. So it can be done by making them sit between the girls or at the extreme ends. So there are a total of 5 places between girls and two places in the extreme ends. So for 5 boys seven places are there to sit so that they satisfy the given condition in the question.
So, the boys can sit in \[{}^{7}{{C}_{5}}\] ways as boys are distinct to each other they can be arranged in 5!. Hence the number of ways for sitting a boy is \[{}^{7}{{C}_{5}}\times 5!\].
Now as we know that there were a total of 6 girls and 5 boys which sum up to 11 people so we can say that they can arrange themselves in 11! ways.
So the total number of outcomes for this situation is 11! And total number of favorable outcomes are \[{}^{6}{{P}_{6}}\times {}^{7}{{C}_{5}}\times 5!\].
So the probability =$\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
So we get,
\[\Rightarrow \dfrac{{}^{6}{{P}_{6}}\times {}^{7}{{C}_{5}}\times 5!}{11!}\]
Here we will use formula,
\[\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
And \[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]
So we can write the expression as,
\[\Rightarrow \dfrac{\dfrac{6!}{0!}\times \dfrac{7!}{\left( 7-5 \right)!\times 5!}\times 5!}{11!}\]
\[\Rightarrow \dfrac{6!\times \dfrac{7!}{2!\times 5!}\times 5!}{11!}\]
Now on the calculation, we get,
\[\Rightarrow \dfrac{6!\times 7!}{2!\times 11!}\]
Hence the probability that no two boys sit together is \[\dfrac{6!7!}{2!11!}\].
Hence the correct option is (C) .

Note: Students generally make the mistakes while finding the number of places where boys have to be seated, they took it as 5 instead of 7 as they left out places that were in the extremities of the row. The student can make the mistake and perform calculations as per this and might end up with the answer as \[\dfrac{7!5!}{2!11!}\] which is option c and loses marks in the exams.