General Math Concepts Review Flashcards
What is the FCP? (Fundamental Counting Principle)
As a general rule, if you have to arrange n different items in order, the total number of orders is the product of the n and every positive integer less than n
ex) 6 different
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
How do you solve this problem?
- 25 employees
- 3 person committee chosen (facilitator, rep, and secretary)
- How many possible committees possible?
25 employees
25 * 24 * 23 = 13800
How do you handle restrictions with the FCP?
With restrictions, break down restrictions into stages with the most restrictive stage coming first, then the next most restrictive, until the only unrestricted stages are left
7 different children: A, B, C, D, E, F, G
7 chairs
child D must be in the middle chair and child G must be next to child D. How many different orders exist?
How do you approach this problem?
Start with the most restrictive stage of child D must be in the middle; list all the ways child D is in the middle. There are two different ways the child D is in the middle and child G is adjacent
A, B, C, D, G, F, E
A, B, G, D, C, F, E
The rest of the kids can be arranged in many ways around them:
5! = 5 * 4 * 3 * 2 * 1 = 120
For the 2 arrangements above, the other 5 can be arranged in 120 different ways each for each of the two
thus:
120 + 120 = 240 total ways to order
What is counting what you don’t want and when should you use it?
n! = total # of arrangements R = # of arrangements that obey the restriction Q = # of arrangements that do not obey
If calculating Q arrangements for a problem is far easier or faster then its better to calculate Q and then subtract from the total to get the answer
Also whenever you see the word “not” in the restriction description then this can be a clue
How do you account for identical items?
You use the mississppi rule:
N = n! / b!
n items
b identical items
For more than one set of identical items:
N = n! / (b!)(c!)(d!)
How do you eliminate repitition in a counting problem?
Understand the same thing can be counted more than once
We eliminate repititions by dividing the total combinations/permutations by the total number of times each arrangement is repeated
What is a combination? What is the equation?
A small group drawn from a large group
n C r = n! / r!(n-r)!
n is total count
r is the group size
What is nCr also equal to? Why?
nC(n-r)
Starting from a pool of n, there must be a group of (n-r) for each combination of r
When do you use combination or when do you know to use it?
When order does not matter!
Typically if each position in the group is the same and there is no difference between positions
Try to understand different orders and see if orders matter for a problem
How can Pascal’s triangle be utilized
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
ex) 7th row of Pascal's triangle 1 7 21 35 35 21 7 1 7C0 = 1 7C1 = 7 7C2 = 21 7C3 = 35 7C4 = 35 ...
What do you do if order matters?
Then its permutations! Use the FCP for it. # permutations = n! where n is the total
How do you handle negative exponents?
B^-m = 1 / B^m
The reason for the above is the following:
b^m / b^n = b^m-n
If m-n = negative number then the following happens
ex)
b^3 / b^7 = b * b * b / b * b * b * b * b * b * b = 1 / b * b * b * b = 1 / b^4
thus b^-4 = 1 / b^4
