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 repeat

Here we select k element groups from n elements, regardless of the order, and the elements can be repeated. k is logically greater than n (otherwise, we would get ordinary combinations). Their 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 is equal to the number of locations of n − 1 separators on n-1 + k places. A typical example is: we go to the store to buy 6 chocolates. They offer only 3 species. How many options do we have? k = 6, n = 3.

Foundation of combinatorics in word problems

Cinema
How many ways can 11 free tickets to the premiere of "Jáchyme throw it in the machine" be divided between 6 pensioners?