3 ms·
If you calculate the number of permutations (order matters), you will double-count a lot of combinations (order does not matter) with the same elements that onl
by _Microft 3y ago
If you calculate the number of permutations (order matters), you will double-count a lot of combinations (order does not matter) with the same elements that only differ by the order of the elements. How many times does that happen? That depends on the number of elements, r. So in how many ways can these elements be arranged? r! To undo this double-counting, you divide nPr by r!.
I feel the situation is much clearer if you actually write nCr and nPr out.
Edit: the Wikipedia articles on combination and permutation are also helpful.
- vector_spaces 3y agoThank you, I think I got it now The idea is that first we choose an r-combination. Once we've chosen a combination of r things, we have to choose a way to order them. Since each r-combinarion has r items, there are r! choices in this second step. This gives the number of r-permutations. Dividing through by r! gives the binomial coefficient n choose r.