JEE PYQ: Permutation And Combination Question 39
Question 39 - 2019 (12 Apr Shift 2)
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:
(1) 28
(2) 27
(3) 25
(4) 24
Show Answer
Answer: (3)
Solution
${}^5C_1 \cdot {}^nC_2 + {}^5C_2 \cdot {}^nC_1 = 1750$. $\frac{5n(n-1)}{2} + 10n = 1750 \Rightarrow 5n^2 - 5n + 20n = 3500 \Rightarrow 5n^2 + 15n - 3500 = 0… $ Actually: $\frac{(n+5)!}{3!(n+2)!} - {}^5C_3 - {}^nC_3 = 1750$. Simpler: $5 \cdot \frac{n(n-1)}{2} + 10n = 1750 \Rightarrow \frac{5n^2 - 5n + 20n}{2} = 1750 \Rightarrow n^2 + 3n - 700 = 0 \Rightarrow n = 25$.
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