Method 2: Find x and y if $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\cdots+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$

If $x,\,y$ are positive integers such that $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}+\dfrac{1}{11!11!}+\dfrac{1}{13!9!}+\dfrac{1}{15!7!}+\dfrac{1}{17!5!}+\dfrac{1}{19!3!}+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$. Find $x,\,y$.

Method 2:

Probability: Coin Tossing

A while back I helped a student with a probability problem, and I took the problem, generalized it a bit, and wish to post it here. Here is the problem:

A coin has the probability $p$ of turning up heads when tossed. Suppose we toss the coin $2n$ times, where $n$ is a natural number. Compute the probability that the total number of heads is even.

Vietnamese Mathematical Olympiad (Trigonometric) Problem of 1962

Solve the equation $\sin^6 x+\cos^6 x=\dfrac{1}{4}$.

This is one of the brilliant Mathematics Olympiad Contest Problems because we can show to the students how there are plenty of ways to attacking a good problem and how one approach is different from the other and how heuristic skill enable us to find solution quickly that save us time for more challenging problems!