Post on 09-Apr-2018
8/7/2019 Counting without repetition
1/4
8/7/2019 Counting without repetition
2/4
n!
C(n,p) = ------------
p!.(n-p)!
Properties of combinations
With the last formula it is easy to prove thata) C(n,p) = C(n,n-p)
b) C(n,p) = C(n-1,p) + C(n-1,p-1) (Pascal's formula)
Binomial coefficient
The number C(n,p) is called a binomial coefficient and this is written as
n
( )
p
Binomial theorem
We'll prove that
(a + b)n = an + C(n,1)an-1 b + C(n,2)an-2 b2 + C(n,3)an-3 b3 + ...
+ C(n,n) bn
To prove this theorem we use mathematical induction.It is easy to verify that the theorem holds for n = 2 .
Now, assume it holds for n = k. We'll show it holds for n= k+1.
(a + b)k+1 = (a + b).(a + b)k
=(a+b).(ak + C(k,1)ak-1 b + C(k,2)ak-2 b2 + C(k,3)ak-3 b3 + ... C(k,k)bk )
= ak+1 + C(k,1)ak b + C(k,2)ak-1 b2 + C(k,3)ak-2 b3 + ... C(k,k)a bk +
ak b + C(k,1)ak-1 b2 + C(k,2)ak-2 b3 + ...+ C(k,k-1)a bk + C(k,k) bk+1
Since C(k,k)= C(k+1,k+1)= 1 and appealing on Pascal's formula
C(n,p) = C(n-1,p) + C(n-1,p-1) , we find
= ak+1 + C(k+1,1)ak b + C(k+1,2)ak-1 b2 + C(k+1,3)ak-2 b3 + ...
+ C(k+1,k+1) bk+1
This proves the theorem.
Counting with repetition
8/7/2019 Counting without repetition
3/4
Variations with repetition
Take a set A of n different elements. Point p elements in a specific order.Each such choice is called a variation with repetition of n elements choose p. Well, there are n
possibilities to point each element.
The total number of variations with repetition of n elements choose p is (np
).
We write
V'(n,p) = np
Permutations with repetition
Example :Take 3 red marbles, 2 blue marbles, and 5 yellow ones. Place this marbles in a specific order.
We call this a permutations with repetition of the marbles. Now, we'll calculate the number ofsuch permutations.
Take 10 numbered compartments. First place the three red marbles in a compartment. This givesC(10,3) possibilities. Then place the 2 blue marbles. Now there are C(10-3,2) possibilities. At
last, we place the 5 yellow marbles. Now there are C(10-3-2,5) possibilities.
The total number of possibilities are
C(10,3).C(10-3,2).C(10-3-2,5)
10! 7! 5!
= -------.-------. ------
3!.7! 2!.5! 5!.0!
10!
= --------------
3! . 2! . 5!
This method can easily be generalized.
Combinations with repetition
Take n different and ordered elements.
Point p elements one after another and order these elements in the same order as the givenelements.
The result is called a combination with repetition of n elements choose p.
We write the number of all combinations with repetition of n elements choose p as C'(n,p).
Example: A = (a,b,c,d,e) and p = 6
Then (a, a, b, d, d, d) ; (b, b, b, c, d, e) ; (c, c, c, c, c, c)are combinations with repetition of 5 elements choose 6.
We can represent such combination by means of a symbol with points and slashes.
8/7/2019 Counting without repetition
4/4
(a,a,b,d,d,d) .././/.../
(b,b,b,c,d,e) /.../././.
(c;c;c;c;c;c) //......//
Each symbol consists of 10 places with exactly 6 points and four slashes. With each such
combination there is just one symbol and with each symbol corresponds just one such
combination.
We can build a symbol putting exactly 6 points in 10 places.
Afterwards, the spare places are filled with slashes.
This can be done in C(10,6) ways.
So, C'(5,6) = C(5+6-1,6)
Generalizing you can prove that C'(n,p) = C(n+p-1,p)