Combinations with repetition

The calculator finds the number of combinations of the k-th class from n elements with repetition. A combination with repetition of k objects from n is a way of selecting k objects from a list of n. The order of selection does not matter and each object can be selected more than once (repeated).

Calculation:

C_{k}^{'} (n) = (k n + k - 1) n = 10 k = 4 C_{4}^{'} (10) = C_{4} (10 + 4 - 1) = C_{4} (13) = (4 1 3) = \frac{1 3 !}{4 ! ( 1 3 - 4 ) !} = \frac{1 3 \cdot 1 2 \cdot 1 1 \cdot 1 0}{4 \cdot 3 \cdot 2 \cdot 1} = 715

The number of combinations with repetition: 715

A bit of theory - the foundation of combinatorics

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.

Combinations with repetition

Calculation:

The number of combinations with repetition: 715

A bit of theory - the foundation of combinatorics

Combinations with repetition

Foundation of combinatorics in word problems