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Calculate the binomial coefficients (a) \(\left(\begin{array}{l}5 \\ 2\end{array}\right)\) (b) \(\left(\begin{array}{r}-2 \\ 3\end{array}\right)\) (c) \(\left(\begin{array}{c}1 / 2 \\ 3\end{array}\right)\) (d) \(\left(\begin{array}{c}-1 / 2 \\ 4\end{array}\right)\)

Short Answer

Expert verified
(a) 10, (b) 0, (c) -1/32, (d) 105/128.

Step by step solution

01

Understanding Binomial Coefficients

Binomial coefficients are denoted as \(\binom{n}{k}\) and represent the number of ways to choose \(k\) elements from a set of \(n\) elements without regard to order. The formula for calculating a binomial coefficient is \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) where \(!\) denotes factorial.
02

Calculate \(\binom{5}{2}\)

Using the formula, we calculate \(\binom{5}{2}\) as follows: \[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4 \times 3!}{2! \times 3!} = \frac{5 \times 4}{2 \times 1} = 10 \] Thus, \(\binom{5}{2} = 10\).
03

Calculate \(\binom{-2}{3}\)

For negative integers, the binomial coefficient \(\binom{n}{k}\) is zero if \(k\) is greater than \(n\) or if either \(n\) or \(k\) is negative. Here, \(n = -2\) and \(k = 3\), so we directly have \(\binom{-2}{3} = 0\).
04

Understanding General Binomial Coefficient for Non-regular Cases

For fractional or general binomial coefficients, the following function is used:\[ \binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!} \] Note that \(n\) can be any real number and \(k\) is a non-negative integer.
05

Calculate \(\binom{1/2}{3}\)

Using the formula for general cases:\[ \binom{1/2}{3} = \frac{(1/2)(1/2-1)(1/2-2)}{3!} = \frac{(1/2)(-1/2)(-3/2)}{6} = \frac{-3/16}{6} = -\frac{1}{32} \] Thus, \(\binom{1/2}{3} = -\frac{1}{32}\).
06

Calculate \(\binom{-1/2}{4}\)

Using the formula for general cases:\[ \binom{-1/2}{4} = \frac{(-1/2)(-3/2)(-5/2)(-7/2)}{4!} = \frac{105}{128} \] Thus, \(\binom{-1/2}{4} = \frac{105}{128}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is an area of mathematics focused on counting, arrangement, and combination of objects. It plays a crucial role in probability and statistics, as it's all about finding the number of ways to arrange or choose items from a larger set. An important part of combinatorics is understanding binomial coefficients, which help in calculating combinations. For instance, when you want to choose a team of 2 out of 5 people, you would use the binomial coefficient \( \binom{5}{2} \), representing the number of combinations of 5 items taken 2 at a time.

In general, combinatorics involves various problems like permutations, combinations, and the study of finite structures. It's widely applicable in computer science, cryptography, and algorithm design.
  • A permutation considers arranging items in a specific order.
  • A combination is about selecting items without regard to order.
  • Combinatorial structures include graphs, design theory, and coding theory.
Understanding these principles is essential for solving problems involving limited resources or discrete structures.
Factorials
Factorials represent the product of an integer and all the positive integers below it. Notated as \( n! \), it's a way to express the number of ways to arrange \( n \) distinct items into a sequence. For example, \( 5! \) equals \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \), indicating 120 different permutations of five items. Factorials are fundamental in computing binomial coefficients and many other mathematical formulas. They simplify calculations by streamlining the multiplication of sequences of decreasing numbers.

The concept of factorials extends beyond whole numbers in advanced math, becoming part of the Gamma function, which generalizes factorials to complex numbers. This extension keeps the properties of factorials useful and applicable in broader situations.
  • 0! is defined as 1, which is a useful identity in many mathematical contexts.
  • Factorials grow rapidly; hence, they're often simplified in equations for computational efficiency.
Negative Integers
When dealing with negative integers in combinatorics, unique rules apply. For binomial coefficients, if the integer part \( n \) is negative and less than \( k \), then \( \binom{n}{k} = 0 \). This happens because you cannot choose more items than are available in a set, aligning with real-world logic. Consider the case of \( \binom{-2}{3} \): choosing 3 items from a set of 2 simply isn't feasible, hence the coefficient is zero.

Negative integers also appear in general cases where \( n \) and \( k \) may not be typical whole numbers. We consider their impacts when calculating binomial coefficients involving negative or fractional values, applying the extended formulas for these instances.
  • Negative integers require careful treatment in equations to ensure valid results.
  • They commonly act as places where counting traditionally fails, warranting special rules and conditions.
Understanding these concepts ensures correct application in mathematical models and problem-solving situations.

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Most popular questions from this chapter

Consider the following puzzle: how many single, loose, smooth \(30 \mathrm{~cm}\) bricks are necessary to form a single leaning pile with no part of the bottom brick under the top brick? Begin by considering. a pile of 2 bricks. The top brick cannot project further than \(15 \mathrm{~cm}\) without collapse. Then consider a pile of 3 bricks. Show that the top one cannot project further than \(15 \mathrm{~cm}\) beyond the second one and that the second one cannot project further than \(7.5 \mathrm{~cm}\) beyond the bottom brick (so that the maximum total lean is \(\left.\left(\frac{1}{2}+\frac{1}{4}\right) 30 \mathrm{~cm}\right)\). Show that the maximum total lean for a pile of 4 bricks is \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}\right) 30 \mathrm{~cm}\) and deduce that for a pile of \(n\) bricks it is \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{2 n+2}\right) 30 \mathrm{~cm}\). Hence solve the puzzle.

Obtain the general solutions of (a) \(Y_{n+2}-7 Y_{n+1}+10 Y_{n}=0\) (b) \(u_{n+2}-u_{n+1}-6 u_{n}=0\) (c) \(25 T_{n+2}=-T_{n}\) (d) \(p_{n+2}-5 p_{n+1}=5\left(p_{n+1}-5 p_{n}\right)\) (e) \(2 E_{n+2}=E_{n+1}+E_{n}\)

Determine the error bound and relative error bound for \(x\), where (a) \(x=35 \mathrm{~min} \pm 5 \mathrm{~s}\) (b) \(x=35 \mathrm{~min} \pm 4 \%\) (c) \(x=0.58\) and \(x\) is correctly rounded to \(2 \mathrm{dp}\).

Expand out (a) and (b) and rationalize (c) to (e). (a) \((3 \sqrt{2}-2 \sqrt{3})^{2}\) (b) \((\sqrt{5}+7 \sqrt{3})(2 \sqrt{5}-3 \sqrt{3})\) (c) \(\frac{4+3 \sqrt{2}}{5+\sqrt{2}}\) (d) \(\frac{\sqrt{3}+\sqrt{2}}{2-\sqrt{3}}\) (e) \(\frac{1}{1+\sqrt{2}-\sqrt{3}}\)

Using exact arithmetic, compute the values of the expressions below. Assuming that all the numbers given are correctly rounded, find absolute and relative error bounds for each term in the expressions and for your answers. Give the answers as correctly rounded numbers. (a) \(1.316-5.713+8.010\) (b) \(2.51 \times 1.01\) (c) \(19.61+21.53-18.67\)

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