# Combination enumeration

Often we are interested in
determining the number of groups

*r*different objects that can be formed from a number of*n*objects. For example, how many groups of 4 different objects can be taken from 5 objects**, and***A, B, C, D***?. To answer this, reason as follows. Because there are 5 ways to choose the first object, 4 ways to choose the second object, and 3 ways to select the third object, then there are all***E***5 x 4 x 3**ways to select group 3 objects if the order of the selected objects is considered. However, because each group of 3 objects, say gurp consisting of objects**A, B**, and**C**, will be enumerated 6 times (meaning all ABC permutations,**ACB, BAC, BCA, CAB**, and**CBA**will be enumerated if the order of objects- this object is noticed, so the number of groups that can be formed is
In general, because

*n (n-1) ... (n-r-1)*states the number of different ways to select a group of*r*objects from*n*available objects if the order is observed and because each group*r*objects will be counted*r!*times in this enumeration, the number of groups*r*different objects that can be formed from a number of n objects is
So denotes
the number of groups

*r*different objects that can be taken from a number of objects if the order in which they are taken is not observed.**Example Combination Problem 1**

A committee of 3 people
will be formed from a total of 20 people. How many committees can be formed?

**Example Combination Problem 2**

From a group of 5 men and
7 women, how many groups of 5 people can be formed consisting of 2 men and 3
women?

Solution: Because there is how to
choose 2 men and how to choose 3 women, then all of them are there

group consisting of 2 men
and 3 women.

A useful combinatorial
identity is

**Equation 1**can be proven by analysis or through the following combinatorial arguments. Suppose we have a group of

*n*objects and focus on any of these objects, name objects number 1. Now there is a combination of size

*r*containing object number 1 (because of this combination this is formed by taking

*r-1*from the remaining

*n-1*objects). In addition, there are combinations of size

*r*that do not contain objects number 1. Because everything is there a combination of size r, then Equation 1 is proven.

Two proofs for this binom
theorem will be given. The first is evidence through mathematical induction,
while the second is evidence through combination arguments.

EVIDENCE OF THEORY OF
BINOM THROUGH INDUCTION: If

*n = 1*,**Equation 2**will be reduced to
Suppose

**Equation 2**is true for*n-1*. Furthermore,
By taking

*i = k + 1*in the first number and*i = k*in the second number we get,
In this case the Equation before the last equation above is a result of Equation 1. So the above theorem proves by induction

COMBINATORIAL EVIDENCE FOR BINOM THEORIES. Look again,

This product represents
the sum of 2

^{n}terms, each of which is a product of*n*factors. In addition, each of these terms has x_{i}or y_{i}as a factor for every*i = 1,2, ..., n*. For example
Now, how much of this 2

^{n}term has*k*x*and*_{i}*n-k*y*as the factor? Because each term consisting of k x*_{i}*and (*_{i}*nk*) y*matches with the selection of a group*_{i}*k*element of*n*values of x*, x*_{1}*, x*_{2}*, ..., x*_{3}*then there is the fruit of the tribe this is so. So, by taking x*_{m}*= x and y*_{i}*= y,*_{i}*i = 1, 2, ..., n*, it is obtained,**Example Combination Problem 3**

Answer:

**Example of Combination Problem 4**

How many subsets can be formed from a set of n elements?

These results can also be obtained in the following ways. To each element in the set

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