Difference between revisions of "User:Temperal/The Problem Solver's Resource5"

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Introduction \| Other Tips and Tricks \| Methods of Proof \| You are currently viewing page 5.

Combinatorics

This section cover combinatorics, and some binomial/multinomial facts.

Permutations

The factorial of a number $n$ is $n(n-1)(n-2)...(1)$ or also as $\prod_{a=0}^{n-1}(n-a)$ ,and is denoted by $n!$ .

Also, $0!=1$ .

The number of ways of arranging $n$ ordered distinct objects is $n!$ . This is also known as a permutation, and can be notated $\,_{n}P_{r}$ . We can see that this is true because there are $n$ objects which you can place in the first spot; when you've picked one there are $n-1$ objects to pick from for the second, and so on.

Combinations

The number of ways of choosing $r$ objects from a set of $n$ objects without replacement (i.e. you can't pick an object twice) is $\frac{n!}{r!(n-r)!}$ , which is notated as either $\,_{n}C_{r}$ or $\binom{n}{r}$ . If you allow replacement, then it's notated $\,_{n}P_{r}$ and is given by $\frac{n!}{(n-r)!}$ . The reader should be able to deduce simple combinatorial arguments for these.

Binomials and Multinomials

Binomial Theorem

$(x+y)^n=\sum_{r=0}^{n}x^{n-r}y^r$

Multinomial Coefficients

The number of ways of ordering $n$ objects when $r_1$ of them are of one type, $r_2$ of them are of a second type, ... and $r_s$ of them of another type so that $\sum r_i=n$ is $\frac{n!}{r_1!r_2!...r_s!}$

Multinomial Theorem

$(x_1+x_2+x_3...+x_s)^n=\sum \frac{n!}{r_1!r_2!...r_s!} x_1+x_2+x_3...+x_s$ . The summation is taken over all sequences $r_i$ so that $\sum_{i=1}^{s}r_i=n$ .

Balls and Urns

There are ${n-1\choose k-1}$ ways to divide $k$ objects in $n$ groups such that no group is empty and the objects are indistinguishable. If groups can be empty, then it's $\binom{n+k-1}{k-1}$

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 __NOTOC__
-<br /><br />
+{{User:Temperal/testtemplate|page 5}}
-{| style='background:lime;border-width: 5px;border-color: limegreen;border-style: outset;opacity: 0.8;width:840px;height:300px;position:relative;top:10px;'
+==<span style="font-size:20px; color: blue;">Combinatorics</span>==
-|+ <span style="background:aqua; border:1px solid black; opacity: 0.6;font-size:30px;position:relative;bottom:8px;border-width: 5px;border-color:blue;border-style: groove;position:absolute;top:50px;right:155px;width:820px;height:40px;padding:5px;">The Problem Solver's Resource</span>
+This section cover combinatorics, and some binomial/multinomial facts.
-|-
+===Permutations===
-| style="background:lime; border:1px solid black;height:200px;padding:10px;" | {{User:Temperal/testtemplate|page 5}}
+The factorial of a number <math>n</math> is <math>n(n-1)(n-2)...(1)</math> or also as <math>\prod_{a=0}^{n-1}(n-a)</math>,and is denoted by <math>n!</math>.
-==<span style="font-size:20px; color: blue;">Limits</span>==
-This section covers limits and some other precalculus topics.
-===Definition===
-*<math>\displaystyle\lim_{x\to n}f(x)</math> is the value that <math>f(x)</math> approaches as <math>x</math> approaches <math>n</math>.
+Also, <math>0!=1</math>.
-*<math>\displaystyle\lim_{x\uparrow n}f(x)</math> is the value that <math>f(x)</math> approaches as <math>x</math> approaches <math>n</math> from values of <math>x</math> less than <math>n</math>.
+The number of ways of arranging <math>n</math> ordered distinct objects is <math>n!</math>. This is also known as a permutation, and can be notated <math>\,_{n}P_{r}</math>. We can see that this is true because there are <math>n</math> objects which you can place in the first spot; when you've picked one there are <math>n-1</math> objects to pick from for the second, and so on.
-*<math>\displaystyle\lim_{x\downarrow n}f(x)</math> is the value that <math>f(x)</math> approaches as <math>x</math> approaches <math>n</math> from values of <math>x</math> more than <math>n</math>.
+===Combinations===
+The number of ways of choosing <math>r</math> objects from a set of <math>n</math> objects without replacement (i.e. you can't pick an object twice) is <math>\frac{n!}{r!(n-r)!}</math>, which is notated as either <math>\,_{n}C_{r}</math> or <math>\binom{n}{r}</math>. If you allow replacement, then it's notated <math>\,_{n}P_{r}</math> and is given by <math>\frac{n!}{(n-r)!}</math>. The reader should be able to deduce simple combinatorial arguments for these.
-*If <math>\displaystyle\lim_{x\to n}f(x)=f(n)</math>, then <math>f(x)</math> is said to be continuous in <math>n</math>.
+===Binomials and Multinomials===
+====Binomial Theorem====
+<math>(x+y)^n=\sum_{r=0}^{n}x^{n-r}y^r</math>
-==Theorems and Properties==
+====Multinomial Coefficients====
+The number of ways of ordering <math>n</math> objects when <math>r_1</math> of them are of one type, <math>r_2</math> of them are of a second type, ... and <math>r_s</math> of them of another type  so that <math>\sum r_i=n</math> is <math>\frac{n!}{r_1!r_2!...r_s!}</math>
-The statement <math>\displaystyle\lim_{x\to n}f(x)=L</math> is equivalent to: given a positive number <math>\epsilon</math>, there is a positive number <math>\gamma</math> such that <math>0<|x-n|<\gamma\Rightarrow |f(x)-L|<\epsilon</math>.
+====Multinomial Theorem====
+<math>(x_1+x_2+x_3...+x_s)^n=\sum \frac{n!}{r_1!r_2!...r_s!} x_1+x_2+x_3...+x_s</math>. The summation is taken over all sequences <math>r_i</math> so that <math>\sum_{i=1}^{s}r_i=n</math>.
-Let <math>f</math> and <math>g</math> be real functions. Then:
-*<math>\lim(f+g)(x)=\lim f(x)+\lim g(x)</math>
-*<math>\lim(f-g)(x)=\lim f(x)-\lim g(x)</math>
-*<math>\lim(f\cdot g)(x)=\lim f(x)\cdot\lim g(x)</math>
-*<math>\lim(\frac{f}{g})(x)=\frac{\lim f(x)}{\lim g(x)}</math>
-Suppose <math>f(x)</math> is between <math>g(x)</math> and <math>h(x)</math> for all <math>x</math> in the neighborhood of <math>S</math>. If <math>g</math> and <math>h</math> approach some common limit L as <math>x</math> approaches <math>S</math>, then <math>\displaystyle\lim_{x\to S}f(x)=L</math>.
+===Balls and Urns===
+There are <math>{n-1\choose k-1}</math> ways to divide <math>k</math> objects in <math>n</math> groups such that no group is empty and the objects are indistinguishable. If groups can be empty, then it's <math>\binom{n+k-1}{k-1}</math>
 [[User:Temperal/The Problem Solver's Resource4|Back to page 4]] | [[User:Temperal/The Problem Solver's Resource6|Continue to page 6]]
-|}<br /><br />

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