How many six-digit numbers exist, for which each succeeding number is smaller than the previous one?
Why are the equalities \(11^2 = 121\) and \(11^3 = 1331\) similar to the lines of Pascal’s triangle? What is \(11^4\) equal to?
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 \(0 = x = y < z\).
Find \(m\) and \(n\) knowing the relation \(\binom{n+1}{m+1}: \binom{n+1}{m}:\binom{n+1}{m-1} = 5:5:3\).
Which term in the expansion \((1 + \sqrt 3)^{100}\) will be the largest by the Newton binomial formula?
How many four-digit numbers can be made using the numbers 1, 2, 3, 4 and 5, if:
a) no digit is repeated more than once;
b) the repetition of digits is allowed;
c) the numbers should be odd and there should not be any repetition of digits?
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.