Download presentation
Published byJulian Lambert Modified over 7 years ago
1
Combinatorial Principles, Permutations, and Combinations
Section 03 Combinatorial Principles, Permutations, and Combinations
2
Permutations vs. Combinations
Permutations are ordered Combinations are not ordered Therefore, there are more permutations than combinations for given π and π Both apply to combinatorics without replacement Also: remember 0! = 1
3
Permutations ORDER MATTERS
Choosing an ordered subset of size π from a collection of π objects without replacement: πππ= π! πβπ ! Given π objects, of which π 1 are Type 1, π 2 are Type 2, etc, up to π π‘ , the number of ways to order all π objects is: π! π 1 !β π 2 !ββ¦β π π‘ !
4
Combinations ORDER DOES NOT MATTER
Choosing a subset of size π from a collection of π objects without replacement: π π =ππΆπ= π! π!β πβπ ! π π is also called a binomial coefficient
5
Binomial Theorem π π is called the binomial coefficient because it is used in the power series expansion of (1+π‘) π (1+π‘) π = π=0 β π π β π‘ π = 1+ππ‘+ π(πβ1) 2 π‘ 2 + π πβ1 (πβ2) 6 π‘ 3 +β¦ If N is as integer, summations stops at k=N If N is not an integer, series is only valid if -1<t<1 This expansion is useful for understanding the binomial distribution
6
Multinomial Theorem In the power series expansion of ( π‘ 1 + π‘ 2 +β¦+ π‘ π ) π the coefficient of π‘ 1 π 1 β π‘ 2 π 2 ββ¦β π‘ π π is π π 1 π 2 β¦ π π = π! π 1 !β π 2 !ββ¦β π π ! Useful for understanding multinomial distributions (much later)
7
In conclusionβ¦ Ordered? Not ordered? πππ= π! πβπ !
πππ= π! πβπ ! Not ordered? π π =ππΆπ= π! π!β πβπ !
8
Sample Exam #4 An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and an unknown number of blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is Calculate the number of blue balls in the second urn
9
Sample Exam #248 Bowl I contains eight red balls and six blue balls. Bowl II is empty. Four balls are selected at random, without replacement, and transferred from bowl I to bowl II. One ball is then selected at random from bowl II. Calculate the conditional probability that two red balls and two blue balls were transferred from bowl I to bowl II, given that the ball selected from bowl II is blue.
10
Actex, Sec 3 #1, pg 103 A class contains 8 boys and 7 girls. The teacher selects 3 of the children at random and without replacement. Calculate the probability that the number of boys selected exceeds the number of girls selected.
11
Actex, Sec 3, #3, pg 103 A box contains 4 red balls and 6 white balls. A sample of size 3 is drawn without replacement from the box. What is the probability of obtaining 1 red ball and 2 white balls, given that at least 2 of the balls in the sample are white?
12
Actex, Sec 3, #5 A box contains 35 gems, of which 10 are real diamonds and 25 are fake diamonds. Gems are randomly taken out of the box, one at a time without replacement. What is the probability that exactly 2 fakes are selected before the second real diamond is selected?
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.