Factorial notation
In a nutshell
Using Pascal's triangle to expand (x+y)n can be inefficient. It is instead far quicker to use combinations; these often involve many terms which look like m×(m−1)×(m−2)×⋯×2×1; factorial notation provides a more compact way of writing such expressions.
Definition
For any positive whole number n:
n!=n×(n−1)×(n−2)×⋯×3×2×1
You pronounce n! as "n factorial". By definition, 0!=1.
Note: For any positive whole number n, n! equals the number of ways of ordering a collection of n objects - you have n choices for the first object, then n−1 choices for the second, and so on. This gives n×(n−1)×⋯×1=n! orderings.
Curiosity: The "most sensible" answer to the question of how many ways you can order a collection of 0 things is widely considered to be 1, which partly justifies defining 0!=1.
Example 1
Compute 3!
Using the above definition:
3!=3×2×1=6
Therefore, 3!=6.
Choosing r objects from n
The number of ways of choosing a collection of r objects from a larger collection of n objects is written as nCr (or sometimes (rn)). This is pronounced "n choose r", and is defined in the following way:
| | | | A non-negative whole number; the number of objects being chosen from.
| | | A non-negative whole number which is at most n; the number of objects being chosen. | | | Stands for "choose"; nCr is the number of ways of choosing r objects from a collection of n. |
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Example 2
What is 3C1?
3C1 denotes the number of ways of choosing 1 object from a collection of 3 objects.
Labelling these objects from 1 to 3, you have 3 choices - either pick object 1, or object 2, or object 3.
Therefore, 3C1=3.
Computing nCr
Imagine choosing a collection of r objects from a larger collection of n objects one by one. You have n choices for the first object, n−1 for the second and so on - until you arrive at n−r+1 choices for the rth object. This gives n×(n−1)×(n−2)×⋯×(n−r+1) choices. More compactly,
n×(n−1)×⋯×(n−r+2)×(n−r+1)=(n−r)!n!
However, you could have chosen those same r objects in any order. There are r! ways to order any collection of r objects, so the above formula has overcounted by a factor of r! Therefore:
nCr=r!(n−r)!n!
Note: Most calculators come with buttons both for factorials and for inputting nCr directly, making values of nCr quick to compute if you have one at hand.
Example 3
What is 6C2?
From the formula dervied above, you have that:
6C2=2!(6−2)!6!=2!×4!6!
Compute:
2!=2×1=2
4!=4×3×2×1=24
6!=6×5×4×3×2×1=720
Substitute these back in:
2!×4!6!=2×24720=15
Therefore, 6C2=15.
Connection to Pascal's triangle
The following two statements about values of nCr turns out to hold true for any positive whole numbers r and n with r<n:
n−1Cr−1+ n−1Cr= nCr
n−1C0= n−1Cn−1=1
These are the very same rules that you use to compute the nth row of Pascal's triangle from the row above; using this, it is possible to show that the rth entry of the nth row of Pascal's triangle is in fact equal to n−1Cr−1.
Example 4
Without using a calculator, work out the 7th entry of the 10th row of Pascal's triangle.
The 7th entry of the 10th row of Pascal's triangle is equal to:
10−1C7−1= 9C6
Substitute 9 and 6 into the formula for nCr:
9C6=6!(9−3)!9!=3!×6!9!
Simplify the expression:
3!×6!9!=3×2×19×8×7=3×4×7=84
Therefore the 7th entry of the 10th row of Pascal's triangle is 84.
