Factorials
N! = N * (N-1) * (N-2) * (N-3) * ... * 3 * 2 * 1 :where N is a positive integer. In many math textbooks, factorial is introduced as the number of ways that an original set can be ordered. For example, if an original set contains 3 items {A, B, C}, then the number of ways that you can order these three items (set elements) is 3! = 3*2*1 = 6 ({A,B,C}, {A,C,B}, {B,A,C}, {B,C,A}, {C,A,B}, {C,B,A}). Note: 0! is defined to equal 1.
| Number |
Factorial
|
... ... ... ... ... | Number |
Factorial
|
|
|
1
|
|
39,916,800
|
|
|
|
1
|
|
479,001,600
|
|
|
|
2
|
|
6,227,020,800
|
|
|
|
6
|
|
87,178,291,200
|
|
|
|
24
|
|
1,307,674,368,000
|
|
|
|
120
|
|
20,922,789,888,000
|
|
|
|
720
|
|
355,687,428,096,000
|
|
|
|
5040
|
|
6,402,373,705,728,000
|
|
|
|
40,320
|
|
121,645,100,408,832,000
|
|
|
|
362,880
|
|
2,432,902,008,176,640,000
|
|
|
|
3,628,800
|
|
BIG!!!!
|
Permutations: If "M" denotes the number of permutations of "N" things taken "P" at a time, then M = N * (N-1) * (N-2) * ... * (N-P+1) = N!/P! (where "!" represents factorial).
For example, if asked how many different ways that one could select a subset of size 3 from an original set of size 5, the answer is (5!)/(2! 3!) = 10. Original set of size five:{A,B,C,D,E} and distinctly different subsets of size 3: {A,B,C}, {A,B,D}, {A,B,E}, {A,C,D}, {A,C,E}, {A,D,E}, {B,C,D}, {B,D,E}, {B,C,E}, and {C,D,E}. Note: Duplicate listings that are NOT distinct ... for example {C,A,B}... are NOT repeated nor listed.
It turns out that the numbers generated by the Binomial Coefficient are identical to those in Pascal's Triangle ...
| K=0 | K=1 | K=2 | K=3 | K=4 | K=5 | K=6 | |
| N=0 | 1 | NA | NA | NA | NA | NA | NA |
| N=1 | 1 | 1 | NA | NA | NA | NA | NA |
| N=2 | 1 | 2 | 1 | NA | NA | NA | NA |
| N=3 | 1 | 3 | 3 | 1 | NA | NA | NA |
| N=4 | 1 | 4 | 6 | 4 | 1 | NA | NA |
| N=5 | 1 | 5 | 10 | 10 | 5 | 1 | NA |
| N=6 | 1 | 6 | 15 | 20 | 15 | 6 | 1 |
where "N" is the row number in Pascal's Triangle (starting a count with row 0), and "K" is the position in the Pascal's Triangle row (starting a count with position 0).
( A + B )0 = 1
( A + B )1 = 1A + 1B
( A + B )2 = 1A2 + 2AB + 1B2
( A + B )3 = 1A3 + 3A2B + 3AB2 + 1B3
( A + B)4 = 1A4 + 4A3B
+ 6A2B2 + 4AB3
+ 1B4