Combination and Permutations Flashcards
Combination
the order does not matter
Permutation
the order does matter
Combination example
- select 5 candies from a dish containing 10 different color candies
- A tycoon has 10 different sports cars in her garage and she will choose three of them to wax. How many different sets of three cars could she choose?
Permutation example
A tycoon has 10 different sports cars in his garage. How many possible ways are there for him to arrange three of the cars in a line?
- It will matter which car goes where
Combination Formula
nCk= n!/ (k!(n-k)!)
Box and fill method for Combination problem
let each box represent a specific decision that must be made. Then divide the product of all the numbers in the boxes by the factorial of the number of boxes that have numbers inside.
A coach must select 3 softball players from a pool of 7 possible players to play for his team. How many different ways are available to the coach in selecting these 3 players?
3!
= 35
If there are m ways to perform one task and n ways to perform another task and of the tasks are independent
m*n possible ways to perform another task
i. e We are making breakfast one morning and that we can choose among 5 types of eggs, 3 types of coffees, and 6 types of pastries. If a breakfast consisting of 1 egg, 1 pastry, and 1 types of coffee, we can then prepare breakfast in one of a total of 536= 90 different ways
i. e: If nickel is flipped five times, how many different head and tail sequences are possible?
- > 2^5= 32
Choosing multiples items from the word “or”
we’re being asked to choose between alternatives
Mutually exclusive
If two event cannot occur together
-> stock market can not go up and down and be unchanged -> not all can happen at the same time
Mutually exclusive -> ways to accomplish it
If there are x ways to accomplish event A and y ways to accomplish event B and if A and B are mutually exclusive then there are x+y ways to accomplish A or B
Collectively exhaustive
those events represent all of the possible outcomes of a scenario
The total ways in which collectively exhaustive scenario can occur
The number of ways in which A can occur + number of ways in which b can occur
Some items can never be together in the same group
The total 3 person club configuration that can be made
number of 3 person club configuration in which both people are together in the club + number of 3 person club configuration in which both people are not together in the club
three people are to be selected from a pool of five people to form a writing club. If two of the people can never be together in the club, in how many different ways can the club be formed?
No restrictions
5C3= 10
Two people are together
3C1= 3
10-3= 7 people
number of 3 person club configuration in which both people are not together in the club
The total 3 person club configuration that can be made - number of 3 person club configuration in which both people are together in the club
number of ways in which at least one of the x items must be chosen from y items
total number of ways to choose x items from y items - the number of ways to choose none of the x items from y items
An international rugby team composed of four people must be organized out of a pool of five players from turkey and six players from Iran, If a team must have at least one player from Iran, in how many different ways can the team be selected?
total # of ways to form team= total- none
11C4= 330
5C4= 5
= 330-5
Dependent Combinations
Two events A and B are dependent if the number of ways in which event B could be selcted depends on which specific way even A is selected
Calculate an unknown number of items in a group
set up a box and fill diagram
->
Permutation
the order of the selection of objects in which the orders of the objects matters
-> In a competition involving ten competitors, one Gold, one sliver and one bronze medal will be awarded. How many different four digit number sequences are possible ?
Permutation formula
nPk= n!/ (n-k)!
Permutation formula using the box method
let each box represent a specific choice that must be made. Multiply the number in each box -> you don’t need to divide by the factorials in the denominator unlike for combination
indistinguishable items in permutation problems
count only the number of distinguishable permutations
-> How many ways are there to arrange SSSSS -> only one way
Permutation formula for indistinguishable items
P= n!/ (r1!)*(r2)!…(rn!)
-> How many ways can A,A,B,B be arranged
P = 4!/ (2!*2!)= 6 ways
Circular arrangement formula
(k-1)!
Permutation anchoring to solve restrictions
first, place the restricted items into their specific spots in the arrangement then handle the additional items that can be arranged in the remaining spots
When some items must be together
link those items together
When two items cannot be next to each other
Two outcomes
1) WHere they stand next to each other
2) When do not stand together
= Total - when they are together
