What is the formula of the number of combination of n objects taken r at a time

The following is a collection of JavaScript for computing permutations and combinations counting with or without repetitions.

Many disciplines and sciences require the answer to the question: How Many? In finite probability theory we need to know how many outcomes there would be for a particular event, and we need to know the total number of outcomes in the sample space.

Combinatorics, also referred to as Combinatorial Mathematics, is the field of mathematics concerned with problems of selection, arrangement, and operation within a finite or discrete system. Its objective is: How to count without counting. Therefore, One of the basic problems of combinatorics is to determine the number of possible configurations of objects of a given type.

You may ask, why combinatorics? If a sample spaces contains a finite set of outcomes, determining the probability of an event often is a counting problem. But often the numbers are just too large to count in the 1, 2, 3, 4 ordinary ways.

A Fundamental Result: If an operation consists of two steps, of which the first can be done in n1ways and for each of these the second can be done in n2 ways, then the entire operation can be done in a total of n1× n2 ways.

This simple rule can be generalized as follow: If an operation consists of k steps, of which the first can be done in n1 ways and for each of these the second step can be done in n2 ways, for each of these the third step can be done in n3 ways and so forth, then the whole operation can be done in n1 × n2 × n3 × n4 ×.. × nk ways.

Numerical Example: A quality control inspector wishes to select one part for inspection from each of four different bins containing 4, 3, 5 and 4 parts respectively. The total number of ways that the parts can be selected is 4×3×5×4 or 240 ways.

Factorial Notation: the notation n! (read as, n factorial) means by definition the product:

n! = (n)(n-1)(n-2)(n-3)...(3)(2)(1). Notice that by convention, 0! = 1, (i.e., 0! º 1) . For example, 6! = 6×5×4×3×2×1 = 720

Permutations versus Combination: A permutation is an arrangement of objects from a set of objects. That is, the objects are chosen from a particular set and listed in a particular order. A combination is a selection of objects from a set of objects, that is objects are chosen from a particular set and listed, but the order in which the objects are listed is immaterial.

The number of ways of lining up k objects at a time from n distinct objects is denoted by n P k, and by the preceding we have:

n P k = (n)(n-1)(n-2)(n-3)......(n-k+1) Therfore, The number of permutations of n distinct objects taken k at a time can be written as:

n P k = n! / (n - k) ! Combinations: There are many problems in which we are interested in determining the number of ways in which k objects can be selected from n distinct objects without regard to the order in which they are selected. Such selections are called combinations or k-sets. It may help to think of combinations as a committee. The key here is without regard for order.

The number of combinations of k objects from a set with n objects is n C k. For example, the combinations of {1,2,3,4} taken k=2 at a time are {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, for a total of 6 = 4! / [(2!)(4-2) !] subsets.

The general formula is:

n C k = n! / [k! (n-k) !].

Permutation with Repetitions: How many different letter arrangements can be formed using the letters P E P P E R?

In general, there are multinomial coefficients:

n! / (n1! n2! n3! ... nr!) different permutations of n objects, of which n1 are alike, n2, are alike, n3 are alike,..... nr are alike. Therefore, the answer is 6! /(3! 2! 1!) = 60 possible arrangements of the letters P E P P E R.

MENU:

1. Permutation of n objects in a group of size k
2. Permutation of n objects in a group of size k, repetitions allowe
3. Combination of n objects in a group of size k
4. Combination of n objects in a group of size k, repetitions allowed

Enter positive integer values for both n and k, and then click on the Calculate.

Chosing subsets its not about permutations and repeated permutations, but it's just about combinations. Your problem is quite easy to understand practically - answer: $3$ ($\{B, R\}, \{R, R\}, \{B, B\}$). Remember that (unordered) sets chosen don't count distinctly if you change order ($\{B, R\} = \{R, B\}$).

To solve this, in general case, with $k$ type of objects, $m$ objects in a subset, $n_m$ objects with at least $m$ occurencies, $n_{m - 1}$ objects with at least $m - 1$ occurencies and so on may not be as easy as understanding a practical idea, as these require analytical solutions (there are combinatorial objects regarding multisets, not simple sets).

Simple is to understand and solve the case with $\alpha$ types of objects with at least $2$ occurences and $\beta$ types of objects with $1$ occurence, chosen in . There are $\binom{\alpha}{2}$ ways to choose pairs of identical objects and $\binom{\alpha + \beta}{2}$ ways to choose pairs of distinct objects. So, our combinatorial object is defined as $\binom{\alpha, \beta}{2} = \binom{\alpha}{2} + \binom{\alpha + \beta}{2}$, a generalisation of standard combinations.

Note that $\binom{n}{r}$ is the binomial coefficient, an expression of combinations. You should read more on these topics to understand the difference between different combinatorial objects like permutations, repetition permutations, combinations, arrangeaments (here order is important!) and nultiset combinatorics.

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