191

Total Ways = $$$\binom{14}{5}$$$. Out of $$$14$$$ places choose $$$5$$$ places for men. Women will follow.
There are total $$$7$$$ pairs of seats. Choose $$$5$$$ out of them. For each pair of seats there are $$$2$$$ ways.
So required ways = $$$\binom{7}{5} \cdot 2^5$$$
Required Probability = $$$\frac{\binom{7}{5} \cdot 2^5}{\binom{14}{5}}= \frac{48}{143}$$$.

607

Two lines can only intersect once.
Maximum number of intersections = $$$\binom{n}{2}$$$ (Choose any two lines and they intersect)
Maximum number of intersections occur when for each intersection point there are only two lines intersecting at that point. Lets label these $$$T2$$$ intersection points.
If there are intersection points where there are $$$x$$$ ($$$x$$$ > $$$2$$$) lines intersecting at some point, then we will lose $$$T2$$$ points.
Amount of point we lost = Amount of points $$$x$$$ lines could have contributed = $$$\binom{x}{2}$$$
So the final answer becomes
$$$\binom{40}{2} - 3\binom{3}{2} - 4\binom{4}{2}- 5\binom{5}{2}- 6\binom{6}{2} = \boxed{607}.$$$

12

Power of $$$x$$$ in $$$n!$$$ = $$$\lfloor \frac{n}{x} \rfloor + \lfloor \frac{n}{x^{2}} \rfloor+ \lfloor \frac{n}{x^3{}} \rfloor + ...$$$ Using this formula we get.. $$$13! = 2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
For a number to be perfect square it prime factorisation should consist of even powers.
Let's write it even and odd powers seperately $$$13! = 2^{10} \cdot 3^{4} \cdot 5^{2} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
Now we want sum of all numbers $$$2^{x} \cdot 3^{y} \cdot 5^{z} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
such that $$$0 \leq x \leq 10 $$$ and $$$0 \leq y \leq 4$$$ and $$$0 \leq z \leq 2$$$ and $$$x, y, z$$$ are even
The resultant number will be $$$\sum\limits_{x} \sum\limits_{y} \sum\limits_{z}2^{x} \cdot 3^{y} \cdot 5^{z} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
which is
$$$(\sum\limits_{x} (2^{x}))\cdot (\sum\limits_{y} (3^{y})) \cdot (\sum\limits_{z} (5^{z})) \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
This can be computed using GP series.
final answer = $$$ 2^{1} \cdot 3^{2} \cdot 5^{1} \cdot 7^{3} \cdot 11^{1} \cdot 13^{4}$$$

944

AOPS solution link

$$$\frac{(n+1)^{2}}{4}$$$ , The Optimal Solution would be a spiral (first along the boundaries then curled inside).

AOPS solution link

2023 AIME and 2023 USAJMO

Math problems that aim to aid competitive programming skills.
Target audience : Experts and below.