Question

In how many ways can 5 boys and 3 girls sit in a row so that no two girls are together?

Answer 1

Hint: Find all the ways in which 5 boys and 3girls can be arranged in a row in the form _B_B_B_B_B_ Now, find out number of ways of arrangement of 5 boys in 5 places and 3 girls in 3 places. Now multiply all this together to get the total number of ways.

Complete step by step answer:
We have been given 5 boys and 3 girls. We have to arrange them in such a way that no 2 girls sit together. Thus, we can say it as,
 \[\_\text{B}\_\text{B}\_\text{B}\_\text{B}\_\text{B}\_\]
Let us take boy as B and girl as G.
Now let us find all the cases of filling 3 girls.
\[\begin{align}
  & \text{GBGBGBBB}~~~~~~~~~~\text{GBBGBBGB}~~~~~~~~~~\text{BGBGBGBB}~~~~~~~~~~\text{BGBBBGBG} \\
 & \text{GBGBBGBB}~~~~~~~~~~\text{GBBGBBBG}~~~~~~~~~~\text{BGBGBBGB}~~~~~~~~~~\text{BBGBGBGB} \\
 & \text{GBGBBBGB}~~~~~~~~~~\text{GBBBGBGB}~~~~~~~~~~\text{BGBGBBBG}~~~~~~~~~~\text{BBGBGBBG} \\
 & \text{GBGBBBBG}~~~~~~~~~~\text{GBBBGBBG}~~~~~~~~~~\text{BGBBGBGB}~~~~~~~~~~\text{BBGBBGBG} \\
 & \text{GBBGBGBB}~~~~~~~~~~\text{GBBBBGBG}~~~~~~~~~~\text{BGBBGBBG}~~~~~~~~~~\text{BBBGBGBG} \\
\end{align}\]
Thus counting all the cases there are 20.
Now we have been given 5 boys. Now these 5 boys can be arranged in 5! Ways.
\[\therefore \text{ Ways to arrange boys}=\text{5}!=\text{5}\times \text{4}\times \text{3}\times \text{2}=\text{12}0\text{ ways}\]
    \[\text{Ways to arrange girls}=\text{3}!=\text{3}\times \text{2}\times \text{1}=\text{6 ways}\]
\[\begin{align}
  & \therefore \text{Total no of ways=}20\times \text{ways to arrange boys }\times \text{ ways to arrange girls} \\
 & \Rightarrow \text{2}0\text{ x 12}0\text{ x 6 }=1\text{44}00\text{ total ways} \\
\end{align}\]
Thus we have arranged 5 boys and 3 girls in 14400 ways in such a way that no 2 girls are together.

Note:
We can also find \[\_\text{B}\_\text{B}\_\text{B}\_\text{B}\_\text{B}\_\]
The number of ways 5 boys can sit : \[{}^{5}{{P}_{5\text{ }}}\text{=}\dfrac{5!}{(5-5)!}\text{=5}\times \text{4}\times \text{3}\times \text{2}\times \text{1}=\text{120 ways}\]
Number of ways in which 3 girls can sit in 6 places
\[\Rightarrow {}^{6}{{P}_{3}}=\dfrac{6!}{(6-3)!}\text{ =6}\times \text{5}\times \text{4}=\text{120 ways}\]
Total number of ways \[\Rightarrow \text{12}0\times \text{12}0=\text{144}00\text{ ways}\].