Combinations without repetition n=100, k=11 result

Find out how many different ways you can choose k items from n items set. With/without repetition, with/without order.

Calculation:

C_{k} (n) = (k n) = \frac{n !}{k ! ( n - k ) !} n = 100 k = 11 C_{11} (100) = (1 1 1 0 0) = \frac{1 0 0 !}{1 1 ! ( 1 0 0 - 1 1 ) !} = 141629804643600

The number of combinations: 141629804643600

141629804643600

A bit of theory - the foundation of combinatorics

Variations

A variation of the k-th class of n elements is an ordered k-element group formed from a set of n elements. The elements are not repeated, and the order of the group's elements matters.

The number of variations can be easily calculated using the combinatorial rule of product. For example, if we have a set of n = 5 numbers {1, 2, 3, 4, 5} and we need to make third-class variations, then V(3,5) = 5 × 4 × 3 = 60.

V_{k} (n) = n (n - 1) (n - 2) . . . (n - k + 1) = \frac{n !}{( n - k ) !}

n! is called the factorial of n, which is the product of the first n natural numbers. The factorial notation is clearer and equivalent. For calculations, it is sufficient to use the procedure derived from the combinatorial rule of product.

Permutations

A permutation is an ordered arrangement of all n elements of a set, where each element is used exactly once, the order matters, and no repetition is allowed.

P (n) = n (n - 1) (n - 2) . . . 1 = n!

Example: We have 4 books. In how many ways can we arrange them side by side on a shelf?

Variations with repetition

A variation with repetition of the k-th class of n elements is an ordered k-element group formed from a set of n elements, where elements can be repeated and order matters. A typical example is forming numbers from the digits 2, 3, 4, 5 and counting how many such numbers exist. We calculate the count using the combinatorial rule of product:

V_{k}^{'} (n) = n \cdot n \cdot n \cdot n . . . n = n^{k}

Permutations with repetition

A permutation with repetition is an ordered arrangement of n elements where some elements appear more than once. The repetition of elements reduces the total number of distinct permutations.

P_{k_{1} k_{2} k_{3} . . . k_{m}}^{'} (n) = \frac{n !}{k _{1} ! k _{2} ! k _{3} ! . . . k _{m} !}

A typical example: find how many distinct seven-digit numbers can be formed from the digits 2, 2, 2, 6, 6, 6, 6.

Combinations

A combination of the k-th class of n elements is an unordered k-element group formed from a set of n elements. The elements are not repeated and the order does not matter. In mathematics, such unordered groups are called sets and subsets. The count is called a combination number and is calculated as follows:

C_{k} (n) = (k n) = \frac{n !}{k ! ( n - k ) !}

A typical example: we have 15 students and need to choose 3. How many ways can this be done?

Combinations with repetition

Here we select k-element groups from n elements, regardless of order, where elements can be repeated. k is generally greater than n (otherwise we would obtain ordinary combinations). The count is:

C_{k}^{'} (n) = (k n + k - 1) = \frac{( n + k - 1 ) !}{k ! ( n - 1 ) !}

Explanation of the formula: the number of combinations with repetition equals the number of ways to place n − 1 separators among n − 1 + k positions. A typical example: we go to a store to buy 6 chocolates. There are only 3 types available. How many different selections can we make? k = 6, n = 3.