Given that, in a tournament, there are $43$ junior level and $51$ senior-level participants ( Boys + Girls).
Let the number of girls in junior-level be $ \text{‘G’}. $
And, the number of boys in senior-level be $ \text{‘B’}.$
The number of girl Vs girl matches in junior-level $ = 153$
$ \Rightarrow \;^{\text{G}}c_{2} = 153 $
$ \Rightarrow \frac{\text{G!}}{(\text{G-2})! \; 2!} = 153 $
$ \Rightarrow \frac{\text{G (G-1) (G-2)!}} { \text{(G-2)}! \; 2!} = 153 $
$ \Rightarrow \text{G (G-1)} = 306 $
$ \Rightarrow \text{G}^{2} – \text{G} – 306 = 0 $
$ \Rightarrow \text{G}^{2} – 18 \text{G} + 17 \text{G} – 306 = 0 $
$ \Rightarrow \text{G(G-18)} + 17 \text{(G-18)} = 0 $
$ \Rightarrow \text{(G-18)(G+17)} = 0 $
$ \Rightarrow \boxed{ \text{G} = 18, \; – 17} $
Thus, the number of girls in junior level $ \boxed {\text{G}= 18} $
So, the number of boys in junior level $ = 43 – 18 = 25.$
The number of matches played between a boy and a girl $ = 25 \times 18 = 450 $
The number of boy Vs boy matches in senior-level $ = 276 $
$ \Rightarrow \;^{ \text{B}}c_{2} = 276 $
$ \Rightarrow \frac{\text{B}!}{(\text{B-2})! \; 2!} = 276 $
$ \Rightarrow \frac{\text{B(B-1)(B-2)!}} {(\text{B} – 2)! \; 2!} = 276 $
$ \Rightarrow \text{B}^{2} – \text{B} = 552 $
$ \Rightarrow \text{B}^{2} – \text{B} – 552 = 0 $
$ \Rightarrow \text{B}^{2} – 24\text{B} + 23 \text{B} – 552 = 0 $
$ \Rightarrow \text{B(B-24)} + 23 \text{(B-24)} = 0 $
$ \Rightarrow \text{(B-24)(B+23)} = 0 $
$ \Rightarrow \boxed{\text{B} = 24, \; -23 } $
Thus, the number of boys in senior level $ \boxed{ \text{B} = 24} $
So, the number of girls in senior level $ = 51 – 24 = 27.$
The number of matches played between a boy and a girl $ = 27 \times 24 = 648.$
$\therefore$ The number of matches a boy plays against a girl $ = 450 + 648 = 1098.$
Correct Answer $:1098$
$\textbf{PS:}$ Among a group of $n$ person, number of matches played between them $ = \;^{n}c_{2} $
$\quad = \frac{n!}{(n-2)! \; 2 \times1} $
$\quad = \frac{n(n-1)(n-2)!}{(n-2)! \; 2 \times 1} $
$\quad = \frac{n(n-1)}{2} $