1
Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $$$\frac{m}{n},$$$ where $$$m$$$ and $$$n$$$ are relatively prime positive integers. Find $$$m+n.$$$
SolutionTotal 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 = $$$\frac{\binom{7}{5} \cdot 2^5}{\binom{14}{5}}= \frac{48}{143}$$$.
2
A plane contains $$$40$$$ lines, no $$$2$$$ of which are parallel. Suppose there are $$$3$$$ points where exactly $$$3$$$ lines intersect, $$$4$$$ points where exactly $$$4$$$ lines intersect, $$$5$$$ points where exactly $$$5$$$ lines intersect, $$$6$$$ points where exactly $$$6$$$ lines intersect, and no points where more than $$$6$$$ lines intersect. Find the number of points where exactly $$$2$$$ lines intersect.
SolutionTwo 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
$$$\binom{40}{2} - 3\binom{3}{2} - 4\binom{4}{2}- 5\binom{5}{2}- 6\binom{6}{2} = \boxed{607}.$$$
3
The sum of all positive integers $$$m$$$ for which $$$\tfrac{13!}{m}$$$ is a perfect square can be written as $$$2^{a}3^{b}5^{c}7^{d}11^{e}13^{f}$$$, where $$$a, b, c, d, e,$$$ and $$$f$$$ are positive integers. Find $$$a+b+c+d+e+f$$$.
SolutionPower of $$$x$$$ in $$$n!$$$ = $$$\lfloor{frac{n}{x} + farc{n}{x^{2}} + frac{n}{x^3{}} + ..\rfloor$$$
What is Misa-MathMath problems that aim to aid competitive programming skills.
Target audience : Experts and below.
