Have the $5$ boys stand in a line. This can be done in $5!$ ways. For the moment, add a boy at each end, who will be removed when we're done. Now send the $3$ girls, one at a time, to stand between two boys. The first girl has $6$ choices, the second girl will have $5$ choices, and the third girl will have $4$ choices. This gives a total of
$$5!\cdot6\cdot5\cdot4=14{,}400$$
arrangements. Remove the two extra boys, and have the others sit.
Remark: the fiction of the extra two boys is simply to avoid having to elaborate that each girl can either wedge herself between two boys or else stand next to a boy at one end or the other.
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Example 24 In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together? The seating arrangement would be as follows Hence total number of ways they can sit = 120 × 120 = 14400 ways 5 girls can sit in any of 5 places Number of ways they can sit = 5P5 = 5!/(5 − 5)! = 5!/0! = 5!/1 = 5 × 4 × 3 × 2 × 1 = 120 3 boys can sit at any of 6 places marked Number of ways they can sit = 6P3 = 6!/(6 − 3)! = 6!/3! = (6 × 5 × 4 × 3!)/3! = 120
In how many ways 5 boys and 3 girls can be seated in a row, so that no two girls are together?
5 boys can be seated among themselves in 5P5 = 5! Ways. After this arrangement, we have to arrange the three girls in such a way that in between two girls there atleast one boy.
So the possible places girls can be placed with the × symbol given below.
× B × B × B × B × B ×
∴ There are 6 places to seated by 3 girls which can be done 6P3 ways.
∴ Total number of ways = 5! × 6P3
= 120 × (6 × 5 × 4)
= 120 × 120
= 14400
Is there an error in this question or solution?
Text Solution
Solution : In order that no two girls are together , we must arrange the 5 boys, each denoted by B, as under: <br> X B X B X B X B X B X <br> Here, B denotes the position of a boy and X that of a girl. <br> Number of ways in which 5 boys can be arranged at 5 places <br> `=""^(5)P_(5)=5! =(5xx4xx3xx2xx1)=120.` <br> Number of ways in which 3 girls can be arranged at 6 places (each marked X) `=""^(6)P_(3)=(6xx5xx4)=120.` <br> Hence, by the fundamental principle of multiplication, the required number of ways `=(120xx120)=14400.
Answer
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 \[\Rightarrow \text{12}0\times \text{12}0=\text{144}00\text{ ways}\].