The first lesson for algebra help will be the introduction of two concepts: Combinations and Permutations.  This is basically the first time within the student’s college career that statistical analysis is involved, typically beginning in Algebra 2 or Intermediate Algebra.  The algebra for combinations and permutations are not done with the use of a simple plug-and-play equation. Typically, these problems require an understanding of a word problem to form an algebraic equation.  There is no set and defined formula provided for you, analyzing the data provided must be done to create it.  Here are some examples of the algebra help our experts provide.

A permutation is defined as: The arrangement of variables within a set.  Basically, this could be simplified into “Having a set number of variables and finding the different possibilities for them.”  It is best described through an example-

Lets take the numbers 1278 for example.  How many ways can we create a 3-number combination or PERMUTATE 3 numbers together?

Answer:
For this problem its possible to come up with the solution manually to find all the variations-
127 217 721 821 187 872
172 218 712 812 178 827
128 271 718 871 278 781
182 281 728 817 287 782

This gives us a total of 24 permutations.  Easily done.  But if I said to find the solution for how many 3-number permutations you can contrive out of every number 1 through 9?  Algebra help can be hard to find so we need something to assist us in this problem.  It would take far too long to write out manually, and so a formula is used:

P(n, r) = n!/(n-r)!
Where n= number of variables to work with
r= size of combination trying to be achieved

So, relate back to the previous 1278 example.  The equation would be filled out like so:
P(4, 3)= 4! / (4-3)!
P( 4, 3)= (4 x 3 x 2 x1) / (1)
P(4, 3)= (24/1)
P(4, 3)= 24
Just as before our solution is 24. 

What if the question looked like this:
From a team of 12, find the total number of different arrangements of 4 people.

Answer:
For this type of problem, an adjustment must be done to our equation.  To find the number of combinations of k elements from a set of n, the formula becomes:

C(n, k)=   n! / (k! * (n-k)!)
Where n= Total number of elements
k= Combination amount required

Back to our problem, the proposed equation should look like this:
C(12, 4) = 12! / (4! * (12-4)!)
C(12, 4) = 12x11x10x9x8x7x6x5x4x3x2x1 / (4x3x2x1 * (8*7*6*5*4*3*2*1))
C(12, 4) = 479001600 / (24 x 40320)
C(12, 4)= 479001600 / 967680
C(12, 4)= 495
There are 495 permutations of 4 people arrangements!


But, as previously mentioned there are no true equations for permutations, its all interpretation.  It put it into perspective:
How many combinations of four letter words can we create using the word CELLULAR? 

A: The number of permutations is computed using the formula listed above.  The problem lies in how permutations do not distinguish between CELL and CELL, for the two different L’s. 
For a problem like this the equation again needs to be adjusted, this time for the repeated variable.
For additional algebra help, contact us immediately!