Question

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

Answer 1

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.

& \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{aligned}$$ Thus counting all the cases there are 20. Now we have been given 5 boys. Now these 5 boys can be arranged in 5! Ways. $$\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{aligned} & \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{aligned}$$ **Thus we have arranged 5 boys and 3 girls in 14400 ways in such a way that no 2 girls are together.** **Note:** We can also find $$\\_\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}$$.