A group of students comprises 5 boys and n gi | vedclass

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(D) Total students = $5 + n$.We need to select $3$ students such that there is at least one boy and at least one girl.The possible cases are:Case $1$:

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$25$

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(D) Total students = $5 + n$.We need to select $3$ students such that there is at least one boy and at least one girl.The possible cases are:Case $1$: $1$ boy and $2$ girls: $^5C_1 	imes ^nC_2 = 5 	imes \frac{n(n-1)}{2} = \frac{5n(n-1)}{2}$.Case $2$: $2$ boys and $1$ girl: $^5C_2 	imes ^nC_1 = 10 	imes n = 10n$.Sum of ways = $\frac{5n(n-1)}{2} + 10n = 1750$.Multiply by $2$: $5n(n-1) + 20n = 3500$.Divide by $5$: $n(n-1) + 4n = 700$.$n^2 - n + 4n = 700 \Rightarrow n^2 + 3n - 700 = 0$.Solving the quadratic equation: $(n + 28)(n - 25) = 0$.Since $n$ must be positive,$n = 25$.

$A$ group of students comprises $5$ boys and $n$ girls. If the number of ways,in which a team of $3$ students can be randomly 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

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