On a plane, six points are given so that no three of them lie on the same line. Each pair of points is connected by a blue or red segment.
Prove that among these points three such points can be chosen so that all sides of the triangle formed by them will be of the same colour.
There are 17 carriages in a passenger train. How many ways can you arrange 17 conductors around the carriages if one conductor has to be in each carriage?
How many ways can you choose four people for four different positions, if there are nine candidates for these positions?
There are \(n\) points on the plane. How many lines are there with endpoints at these points?
Prove the validity of the following formula of Newton’s binom \[(x+y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \dots + \binom{n}{n}y^n.\]
Think of a way to finish constructing Pascal’s triangle upward.
Calculate the following sums:
a) \(\binom{5}{0} + 2\binom{5}{1} + 2^2\binom{5}{2} + \dots +2^5\binom{5}{5}\);
b) \(\binom{n}{0} - \binom{n}{1} + \dots + (-1)^n\binom{n}{n}\);
c) \(\binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n}\).
In the expansion of \((x + y)^n\), using the Newton binomial formula, the second term was 240, the third – 720, and the fourth – 1080. Find \(x\), \(y\) and \(n\).
Show that any natural number \(n\) can be uniquely represented in the form \(n = \binom{x}{1} + \binom{y}{2} + \binom{z}{3}\) where \(x, y, z\) are integers such that \(0 \leq x < y < z\), or \(0 = x = y < z\).
Here is a fragment of the table, which is called the Leibniz triangle. Its properties are “analogous in the sense of the opposite” to the properties of Pascal’s triangle. The numbers on the boundary of the triangle are the inverses of consecutive natural numbers. Each number is equal to the sum of two numbers below it. Find the formula that connects the numbers from Pascal’s and Leibniz triangles.