Permutations & Combinations
Grade None
Question:
<p>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</p>
<p style="display:inline">25</p>
<p style="display:inline">28</p>
<p style="display:inline">27</p>
<p style="display:inline">24</p>
Step-by-Step Solution
Key Concept: Identify and sum the mutually exclusive combinations that satisfy the constraints by breaking the team selection into specific counts of boys and girls.
<p>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)<br />
<span class="math-tex">$=\left(^{5} C_{1} \times^{n} C_{2}\right)\left(^{5} C_{2} \times^{n} C_{1}\right)$</span> [given]<br />
<span class="math-tex">$\Rightarrow\left(5 \times \frac{n(n-1)}{2}\right)+\left(\frac{5 \times 4}{2} \times n\right)=1750$</span><br />
<span class="math-tex">$\Rightarrow n(n-1)+4 n=\frac{2}{5} \times 1750 \Rightarrow n^{2}+3 n=2 \times 350$</span><br />
<span class="math-tex">$\Rightarrow n^{2}+3 n-700=0$</span><br />
<span class="math-tex">$\Rightarrow$</span> n<sup>2</sup> + 28n - 25n-700= 0<br />
<span class="math-tex">$\Rightarrow$</span> n(n + 28) -25(n + 28) = 0<br />
<span class="math-tex">$\Rightarrow$</span> (n -25) (n + 28) = 0<br />
<span class="math-tex">$\Rightarrow$</span> n = 25, (-28 (discarded)<br />
Hence n = 25</p>
Correct Answer: A