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

Gertrude 62304
Six boys and six girls (among them Emil, Félix, Gertrude, and Hanka) want to dance. The number of ways they can make six (mixed) couples if Emil does not want to dance with Gertrude and Hanka wants to dance with Felix is?