1. ## Multiset

Figured out, can this be deleted?

2. $\displaystyle \sum\limits_{k = 0}^n {{{k + 1} \choose k}\left( {n + 1 - k} \right)}$
Can you explain this?

3. Plato, that is the answer, but I have no idea how to get that.

4. ^^actually unless I didn't something wrong, that doesn't hold for n=2

5. The actual sum is this $\displaystyle \sum\limits_{k = 0}^n {\left( {k + 1} \right)\left( {\begin{array}{*{20}c} {n + 1} \\ {n - k} \\ \end{array}} \right)}$. You can choose any number form 0 to n letters.

Say n=10; then you may choose 3 a’s and a b along with any six numbers from 1 to 7. In this case, you have chosen 4 letters and 6 numbers. There are 5 ways to pick 4 letters: aaaa, aaab, aabb, abbb, & bbbb. There are 7 ways to pick six numbers from 1 to 7.

Now we sum on all the cases.

6. Plato then n = 2

should give

(a,a) (a,b) (a,1) (a,2) (a,3)
(b,b) (b,1) (b,2) (b,3)
(1,2) (1,3)
(2,3)

Which is twelve 2 - combinations for n =2

but the formula you put says there should be 10.

7. Originally Posted by Jrb599
Plato then n = 2
Which is twelve 2 - combinations for n =2
but the formula you put says there should be 10.
Yes you are correct it should be
$\displaystyle \sum\limits_{k = 0}^n {\left( {k + 1} \right)\left( {\begin{array}{*{20}c} {n + 1} \\ {n - k} \\ \end{array}} \right)}$

8. Plato, sorry to bug you with questions

but how did you start your logic process? Why do you multiple the two terms?

Also how did you know to sum them up?