Permutation and Combination
Permutation and Combination
Allen Star Batch
Grade 11
Question:
In a plane, there are two families of lines $y = x + r, y = -x + r$, where $r \in \{0, 1, 2, 3, 4\}$. The number of squares of diagonals of length $2$ formed by the lines is:
$\binom{2}{3}(4!)$
$\binom{3}{2}(3!)$
$16$
$9$
Step-by-Step Solution
Key Concept: Logarithmic and exponential equations in binomial problems often reduce to quadratic-type equations after substitution.
From $\frac{14}{9}\binom{9}{2}\binom{9}{4} = \left(\binom{9}{5}\right)^2$ with $m=9$, we use the constraint $\binom{9}{15}^{-\log(6-\sqrt{x})} \cdot 5 = \frac{84}{5}$ to obtain $6\sqrt{2}x - x - 10 = 0$, or $x - 6\sqrt{2}x + 10 = 0$. Factoring as $(\sqrt{x}-5\sqrt{2})(\sqrt{x}-\sqrt{2}) = 0$ gives $x = 2$ (rejecting $x = 50$).
Correct Answer: 2,4