8/23/2023 0 Comments Permutation combination formula![]() ![]() It's not so hard to see that each permutation of these circles corresponds to a different way of putting each these $k$ objects into the $n$ cells. ![]() Let the division between the cells be a white circles and the objects black circles, then there would be $(n-1)$ white circles and $k$ black ones. Now here comes the tricky part, we can count the permutations of this set by cleverly assigning circles. Using the same analogy for combinations with replacement we have $k$ objects that we want to distribute into this $n$ cells but now we can put more than one object per cell (hence with replacement) also note that there is no bound on $k$ because if $k>n$ then we can just put more than one object in each cell. $$(.)= \bigcirc \bullet \bullet \bigcirc \bullet (.)\bullet \bigcirc $$ It is easy to see that this corresponds to a combination without replacement because if we represent the occupied cells with a black circle and the empty cells with a white one there would be $k$ black circles in the row and $(n-k)$ white ones in the row, so the permutations of this row is precisley: Like combinations, there are two types of permutations: permutations with repetition, and permutations without repetition. Permutations can be denoted in a number of ways: n P r, n P r, P(n, r), and more. The key here is that due to the fact that there is no replacement there is only one or zero objects in each cell. In cases where the order doesnt matter, we call it a combination instead. A combination without replacement of $k$ objects from $n$ objects would be equivalent to the number of ways in which these $k$ objects can be distributed among the cells with at most one object per cell. Imagine you have $n$ different cells form left to right.
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