![]() 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. The two key formulas are: Permutation Formula A permutation is the choice of r things from a set of n things without replacement and where the order matters. 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. This article describes the formula syntax and usage of the PERMUT function in Microsoft Excel. $$(.)= \bigcirc \bullet \bullet \bigcirc \bullet (.)\bullet \bigcirc $$ There is a way you can calculate permutations using a formula. The value for (r) is the number of letters we use in each sequence. Since we are selecting from 5 different letters in the set, (n5). The number of three-letter word sequences. 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: Use the permutation formula to find the following: The number of four-letter word sequences. ![]() Still we will see that in many cases it is important to have an explicit formula for the solution of a problem. ![]() ![]() Explicit formulas in a) and b) are rarely used in numerical computation because there are much better algorithms, for example, the row operations. The key here is that due to the fact that there is no replacement there is only one or zero objects in each cell. Again, these formulas have sense, because we assume A to be non-singular, so detA 6 0. 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. k3 3 (3 Ds) Finally the formula is n/k1k2k3 Basically, you get the permutations of those 8 letters then rule out those options which are essentially the same. ![]()
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