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In how many ways can you choose r distinct objects from a collection of
n distinct objects ?
For example, suppose you have 3
cards -- ace, king, and queen.
In how many ways can you choose two cards from those three ?
If the order in which the choices are made makes a difference,
then there are six ways to choose two cards:
| ace | king | ||
| king | ace | ||
| ace | queen | ||
| queen | ace | ||
| king | queen | ||
| queen | king |
We say that there are 6 permutations of three things taken two at a time.
But if the order of choosing doesn't matter,
then there are only 3 ways to pick 2 cards.
| ace | king | ||
| ace | queen | ||
| king | queen |
We say that there are 3 combinations of three things taken two at a time.
| order matters | ↔ | the choice is a permutation | |
| order does not matter | ↔ | the choice is a combination |
| The symbol | ( | n | ) |
represents the number of combinations of
n things taken r at a time. Numbers written in this form are called binomial coefficients . |
| r |
|
It can be shown†
fairly easily that |
||||||||
| ( | n | ) |
|
|||||
| r | ||||||||
From the above equation it would be very easy to prove [ we leave the proofs for you ] these two facts:
| Fact 1 | ||||||||
| ( | n | ) | = |
|
||||
| r | ||||||||
| Fact 2 | ||||||||
| ( | n | ) | = | ( | n | ) | ||
| r | n - r | |||||||
In Fact 1, if we re-write the numerator
of the right hand side as
(n - 0) * (n - 1) * (n - 2) * .... * [ n - ( r - 1 ) ],
then
it is easy to see that there are exactly
r factors there.
So, to evaluate n binomial r you simply multiply
n * ( n - 1 ) * .... until
you have written down r factors,
and then you divide that product by the factorial
of r.
| For example, | ( | 52 | ) | equals ( 52 * 51 * 50 * 49 ) / 4! which is 270,725. |
| 4 |
Fact 2 is helpful whenever r is more than half of n.
| For example, | ( | 20 | ) | = | ( | 20 | ) | = | ( 20 * 19 * 18 ) / 3! which is 1140. |
| 17 | 20 - 17 |
If both n and r are fairly small, say no bigger than 10 or so, then you can also evaluate a binomial coefficient by drawing a little picture of the first few rows of Pascal's Triangle and using the fact that n binomial r equals the number in the nth row and rth column of the triangle.
In Fig 1 below we show columns 0 thru 6 and rows 0 thru 6 of Pascal's Triangle.
The rows of Pascal's Triangle are numbered starting from zero at the top. The columns are slanted at about 45 degrees and are numbered from zero starting at the left hand side.
| As an example, | ( | 6 | ) | = 20, which appears in the 3rd column of row 6 in Fig 1. |
| 3 |
To add another row to the triangle you always begin by putting a 1 into column zero and you end by putting a 1 into the last column. For the inner numbers you look at the row above, and between each pair of numbers you put their sum into the row below.
Fig 2 shows 3 consecutive numbers a, b, and c inside some row. The number x in the following row lies between and below a and b and is found by adding a and b together. Between b and c you put their sum y into the next row.
|
|
If we needed row 7, Fig 3 shows what it would look like.
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