##### Contents

Combinatorics

Probability & Statistics

# Counting analysis

### 1. Combinations

```
Let's consider again the following set of digits Set= {1,2,3}

Question1:
How many numbers of two digits can we obtain from
12, 13, 21, 23, 31, and 32. In total, 6 numbers.

For every number with the same digits, we have exactly 2!=2
repeatitions.

Question2:
How many numbers of two identical digits can we obtain from
We have only three : 12, 13, 23

The order of these numbers IS important. The arragement
13 and 31 , for example, are the same.

The question would be presented as: How many arrangements
of two digits can we make from the three elements of the set Set,
when the order is important?

There are 3 manners to arrange two digits among three digits acording
this constraint.

We process as we did for the arragement. We arrange r objects among
n objects with order. Here the arrangement is called comination:

Since we know the number of counts for arragements, we deduce
the one for combinations. For every number with the same digits,
we have  exactly !r repeatitions. We simply divide the number of
arragements by n! to set the number of combinations. It follows:

The number of combinations of r among n objects is
C(r,n) = A(r,n)/r! = n!/(n-r)! r!

C(r,n) gives also the number of r-subsets possible out of a set
of n distinct elements. For example, The 2-subsets of {1,2,3} are
the three pairs {1,2}, {1,3}, {2,3}, so C(2,3)=3.
```

#### 2. Pascal triangle

```
The Pascal triangle can be constructed from the
extantions of the binomials:

(a + b)0 = 1
(a + b)1 = 1 a1 +  1 b1
(a + b)2 = 1 a2 +  2 ab + 1 b1
(a + b)3 = 1 a3 +  3 a2b + 3 a b2 + 1 b3
(a + b)4 = 1 a4 +  4 a3b + 6 a2b2 +
4 ab3 + 1 b4
(a + b)5 = 1 a5 +  ....

(a + b)0 = C(0,0)
(a + b)1 = C(0,1) a1 +  C(1,1) b1
(a + b)2 = C(0,2) a2 +  C(1,2) ab + C(2,2)b1
(a + b)3 = C(0,3) a3 +  C(1,3) a2b + C(2,3) a b2 + C(3,3)b3
(a + b)4 = C(0,4) a4 +  C(1,4) a3b + C(2,4) a2b2 +
C(3,4) ab3 + C(3,4) b34
(a + b)5 = C(0,5)a5 +  .... + C(5,5)b5

That give the following general tringle: which includes
the binomial coefficients

We have the following relationship:

C(r+1,n) = (n-r)C(r,n)/(r+1)

C(r+1,n)  = n!/(n-r-1)!(r+1)! = n!(n-r)/(n-r)!(r)!(r+1) =
(n-r)/(r+1) x n!/(n-r)!(r)! = (n-r)/(r+1) x C(r,n)
```

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