Permutations And Combinations Ques 52
- A group of students comprises of 5 boys and $n$ girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750 , then $n$ is equal to
(2019 Main, 12 April II)
(a) 28
(b) 27
(c) 25
(d) 24
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Answer:
Correct Answer: 52.(c)
Solution:
Formula:
- It is given that a group of students comprises of 5 boys and $n$ girls. The number of ways, in which a team of 3 students can be selected from this group such that each team consists of at least one boy and at least one girls, is $=$ (number of ways selecting one boy and 2 girls) + (number of ways selecting two boys and 1 girl)
$ \begin{aligned} & =\left({ }^{5} C _1 \times{ }^{n} C _2\right)\left({ }^{5} C _2 \times{ }^{n} C _1\right)=1750 \text { [given] } \\ & \Rightarrow \quad ( 5 \times \frac{n(n-1)}{2}) + ( \frac{5 \times 4}{2} \times n ) =1750 \\ & \Rightarrow n(n-1)+4 n=\frac{2}{5} \times 1750 \Rightarrow n^{2}+3 n=2 \times 350 \\ & \Rightarrow n^{2}+3 n-700=0 \Rightarrow n^{2}+28 n-25 n-700=0 \\ & \Rightarrow n(n+28)-25(n+28)=0 \Rightarrow(n+28)(n-25)=0 \\ & \Rightarrow n=25 \quad \quad[\because n \in N] \end{aligned} $
BETA
