**How do combinations with repetition work? - combinations with repetition and restrictions**

I have learned of combinations and permutations. I have everything, if the variation is replaced by the permutation (with and without repetition), combinations without repetition. With repetition, however, is much more difficult. I http://www.mathsisfun.com/combinatorics/combinations-permutations.html Help me with the first three, but the explanation is there for the last has no meaning for me. I have tried to find another site for me, but he had even less sense.

## 1 comment:

Suppose first that you understand the difference between the combinations with and without recurrence.

For example, Article 3 A, B, C, and can choose from 2, without repetition, then you are AB, AC and AC, but not repetition AA, AB, AC, BB, BC and CC.

I understand that you are trying to understand how to determine the n objects, as can a selection of objects with R disordered repetition done in C (n + r-1, R) in different ways.

To see this, look at the problem differently. Instead of n objects X1, X2, ...., Xn,

r selection with repetition allowed them to think I have checked the containers N X1, X2, ...., Xn and chips st To select the object Xi k times instead of chips, I K Xi.

EC. Suppose you have objects X1, X2, X3, and will select up to 5Objects with repetition. A selection would X1, X1, X1, X3, X3.

Or suppose you have marked 5 tokens of 5 asterisk (*). The above is the selection

Bin *** X1, X2 Bin (empty), X3 Bin **

Note that the counting of the selection of the repetition of n objects r is exactly the same as the counting of the cards are placed in r different ways in bucket n.

As a compact notation, rather than have to think, n-weighted 1 Shelves delimiter | 's, then a row of asterisks * ry (| n-1) separators.

EC.

Bin *** X1, X2 Bin (empty), denoted by Bin X3 ** ***||**

X1 Bin (empty), Bin Bin X2 * X3 rated ****, |****|*

Then, by counting the number of the selection of r with a repetition of n objects exactly the sameas counting how many different ways in which we organize an asterisk * ry (n-1) Separators |.

We have a total of R + n (n-1) = n + r-1-objects (stars and R-1 separator), and want as many options as we know, fix it.

Consider n + r-1 slots, and choose r of them, about the asterisk. This can be done in C (n + r-1, R) fashion. Then put the N-1-separators in vacuum (n + r-1)-r = n-1 slots. This can be done in 1 way. At the beginning of the spread, the task can be carried out in C (n + r-1, R) 1 = C (n + r-1, R) manner.

To select the number of possibilities, with the repetition of r objects selected from C n (n + r-1, R).

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