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    Counting principles, including permutations and combinations.

    The binomial theorem: expansion of 𝒂 + 𝒃 𝒏, 𝒏 𝜺 𝑡.

    ● THE PRODUCT RULE

    If there are π‘š different ways of performing an operation and for each of these there are 𝑛 different ways of performing a second independent operation, then there are π‘šπ‘› different ways of performing the two operations in succession. The product principle can be extended to three or more successive operations. The number of different ways of performing an operation is equal to the sum of the different mutually exclusive possibilities.

    ● COUNTING PATHS

    The word π‘Žπ‘›π‘‘ suggests multiplying the possibilities The word π‘œπ‘Ÿ suggests adding the possibilities. If the order doesn't matter, it is a Combination. If the order does matter it is a Permutation.

    ● PERMUTATIONS (order matters)

    A permutation of a group of symbols is any arrangement of those symbols in a definite order.

    ● Permutations of 𝒏 different object : 𝒏!

    Explanation: Assume you have n different symbols and therefore n places to fill in your arrangement. For the first place, there are n different possibilities. For the second place, there are n – 1 possible symbols, … until we saturate all the places. According to the product principle, therefore, we have n (n – 1)(n – 2)(n – 3)β‹―1 different arrangements, or n!

    Wise Advice: If a group of items have to be kept together, treat the items as one object. Remember that there may

    be permutations of the items within this group too. ● Permutations of π’Œ different objects out of 𝒏 different available (no repetition allowed) :

    π‘ƒπ‘˜ 𝑛 =

    𝑛!

    𝑛 βˆ’ π‘˜ ! = 𝑛 βˆ™ 𝑛 βˆ’ 1 βˆ™βˆ™βˆ™ 𝑛 βˆ’ π‘˜ + 1

    Good logic to apply to similar questions straightforward: Suppose we have 10 letters and want to make groups of 4 letters. For four-letter permutations, there are 10 possibilities for the first letter, 9 for the second, 8 for the third, and 7 for the last letter. We can find the total number of different four-letter permutations by multiplying 10 x 9 x 8 x 7 = 5040.

    ● Permutations with repetition of π’Œ different objects out of 𝒏 different available = π’π’Œ

    There are n possibilities for the first choice, THEN there are n possibilities for the second choice, and so on, multiplying each time.)

    ● COMBINATIONS (order doesn’t matters)

    It is the number of ways of choosing π’Œ objects out of 𝒏 available given that β–ͺ The order of the elements does not matter. β–ͺ The elements are not repeated [such as lottery numbers (2,14,15,27,30,33)]

    The easiest way to explain it is to:

    ο‚· assume that the order does matter (ie permutations),

    ο‚· then alter it so the order does not matter. Since the combination does not take into account the order, we have to divide the permutation of the total number of symbols available by the number of redundant possibilities. π’Œ selected objects have a number of redundancies equal to the permutation of the objects π’Œ! (since order doesn’t matter)

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    However, we also need to divide the permutation n! by the permutation of the objects that are not selected, that is to say 𝒏 βˆ’ π’Œ ! .

    𝒏!

    π’Œ! 𝒏 βˆ’ π’Œ !

    ( 𝒏 π’Œ ) ≑ π‘ͺπ’Œ

    𝒏 ≑ π‘ͺπ’Œ 𝒏 =

    𝒏!

    π’Œ! 𝒏 βˆ’ π’Œ ! =

    𝒏 𝒏 βˆ’ 𝟏 𝒏 βˆ’ 𝟐 𝒏 βˆ’ π’Œ + 𝟏

    π’Œ!

    ● Binomial Expansion/Theorem

    π‘Ž + 𝑏 𝑛 = βˆ‘ ( 𝑛 π‘˜ ) π‘Žπ‘›βˆ’π‘˜π‘π‘˜ = π‘Žπ‘› + (

    𝑛 1 ) π‘Žπ‘›βˆ’1𝑏 + β‹―+ (

    𝑛 π‘˜ ) π‘Žπ‘›βˆ’π‘˜π‘π‘˜ + β‹―+ 𝑏𝑛

    𝑛

    π‘˜=0

    ● Binomial Coefficient

    ( 𝑛 π‘˜ ) is the coefficient of the term containing π‘Žπ‘›βˆ’π‘˜π‘π‘˜ in the expansion of π‘Ž + 𝑏 𝑛

    ( 𝑛 π‘˜ ) =

    𝑛 𝑛 βˆ’ 1 𝑛 βˆ’ 2 β‹― 𝑛 βˆ’ π‘˜ + 1

    π‘˜! =

    𝑛!

    π‘˜! 𝑛 βˆ’ π‘˜ ! =

    𝑛!

    𝑛 βˆ’ π‘˜ ! π‘˜! = (

    𝑛 𝑛 βˆ’ π‘˜

    )

    π‘‡β„Žπ‘’ π‘”π‘’π‘›π‘’π‘Ÿπ‘Žπ‘™ π‘‘π‘’π‘Ÿπ‘š π‘œπ‘Ÿ π‘˜ + 1 π‘‘β„Ž π‘‘π‘’π‘Ÿπ‘š 𝑖𝑠: π‘‡π‘˜+1 = ( 𝑛 π‘˜ )π‘Žπ‘›βˆ’π‘˜π‘π‘˜

    The constant term is the term containing no variables.

    When finding the coefficient of π‘₯𝑛 always consider the set of all terms containing π‘₯𝑛

    Probability ο‚· The number of trials is the total number of times the β€œexperiment” is repeated.

    ο‚· The outcomes are the different results possible for one trial of the experiment.

    ο‚· Equally likely outcomes are expected to have equal frequencies.

    ο‚· The sample space, U, is the set of all possible outcomes of an experiment.

    ο‚· And event is the occurrence of one particular outcome.

    𝑃 𝐴 = 𝑛 𝐴

    𝑛 π‘ˆ

    ο‚· Complementary Events

    Two events are described as complementary if they are the only two possible outcomes.

    Two complementary events are mutually exclusive.

    Since an event must either occur or not occur, the probability of the event either occurring or not

    occurring must be 1.

    𝑷 𝑨 + 𝑷 𝑨′ = 𝟏

    Use when you need probability that an event will not happen

    ο‚· Possibility when we are interested in more than one outcome (events are β€œand”, β€œor”, β€œat least”)

    P(A) is the probability of an event A occurring in one trial,

    n(A) is the number of times event A occurs in the sample space n(U) is

    the total number of possible outcomes.

    𝑛 πœ€ 𝑁 ( 𝑛 0 ) ≑ 1 0! ≑ 1

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    ο‚· Combined Events

    βˆͺ π‘’π‘›π‘–π‘œπ‘› ≑ π‘’π‘–π‘‘β„Žπ‘’π‘Ÿ ∩ π‘–π‘›π‘‘π‘’π‘Ÿπ‘ π‘’π‘π‘‘π‘–π‘œπ‘› ≑ π‘π‘œπ‘‘β„Ž/π‘Žπ‘›π‘‘

    Given two events, B and A, the probability of at least one of the two events occurring,

    𝑃 𝐴 βˆͺ 𝐡 = 𝑃 𝐴 + 𝑃 𝐡 βˆ’ 𝑃 𝐴 ∩ 𝐡

    𝐼𝑑 𝑖𝑠 π‘–π‘šπ‘π‘œπ‘Ÿπ‘‘π‘Žπ‘›π‘‘ π‘‘π‘œ π‘˜π‘›π‘œπ‘€ β„Žπ‘œπ‘€ π‘‘π‘œ 𝑔𝑒𝑑 𝑃 𝐴 ∩ 𝐡

    For mutually exclusive events (no possibility that A and B occurring at the same time)

    Turning left and turning right (you can't do both at the same time)

    Tossing a coin: Heads and Tails

    𝑃 𝐴 βˆͺ 𝐡 = 𝑃 𝐴 + 𝑃 𝐡 𝑃 𝐴 ∩ 𝐡 = βˆ…

    For non - mutually exclusive we are going to find conditional probability for

    Independent and Dependent Events

    A bag contains three different kinds of marbles: red, blue and green. You pick the marble twice. Probability of picking up the red one (or any) the second time depends weather you put back the first marble or not.

    β€’ Independent Events: β€’ Dependent Events:

    the probability that one event occurs probability of one event occurring influences in no way affects the probability of the likelihood of the other event the other event occurring. You put the first marble back You don’t put the first marble

    ∎ Conditional Probability:

    Given two events, B and A, the conditional probability of an event A is the probability that the event will occur given the knowledge that an event B has already occurred. This probability is written as (notation for the probability of A given B) P (A|B )

    Probability of the intersection of A and B (both events occur) is: 𝑃 𝐴 ∩ 𝐡 = 𝑃 𝐡 𝑃 𝐴|𝐡

    β€’ Independent Events: β€’ Dependent Events:

    𝑃 𝐴|𝐡 = 𝑃 𝐴 = 𝑃 𝐴|𝐡′ 𝑃 𝐴 ∩ 𝐡 = 𝑃 𝐡 𝑃 𝐴|𝐡

    𝐴 π‘‘π‘œπ‘’π‘  π‘›π‘œπ‘‘ 𝑑𝑒𝑝𝑒𝑛𝑑 π‘œπ‘› 𝐡 π‘›π‘œπ‘Ÿ π‘œπ‘› 𝐡′ 𝑃 𝐴|𝐡 π‘π‘Žπ‘™π‘π‘’π‘™π‘Žπ‘‘π‘’π‘‘ 𝑑𝑒𝑝𝑒𝑛𝑑𝑖𝑛𝑔 π‘œπ‘› π‘‘β„Žπ‘’ 𝑒𝑣𝑒𝑛𝑑 𝐡

    𝑃 𝐴 ∩ 𝐡 = 𝑃 𝐴 𝑃 𝐡 𝑃 𝐴 ∩ 𝐡 = 𝑃 𝐡 𝑃 𝐴|𝐡

    𝑃 𝐴|𝐡 = 𝑃 𝐴 ∩ 𝐡

    𝑃 𝐡

    either A or B or both

    P(A) includes part of B from intersection P(B) includes part of A from intersection

    𝑃 𝐴 ∩