1. ## Combinatorial Math Problem

I've forgotten the methodology for deducing the number of ways to select N groups from M objects. For example, if I had 11 flags, how many ways would there be to put the 11 flags in 5 piles?

The real question I'm working on is:

"How many integer solutions are there to the equation x1 + x2 + x3 + x4 + x5 = 21, if xi >= 2 for i = 1, 2, 3, 4, 5."

I was going to try and think about it as having 5 groups of at least 2 (since each xi much be at least 2) and then distribute the remaining 11 groups of 1 between the 5 groups.

The number of non-negative integer solutions to $x_1 + x_2 + x_3 + x_4 + x_5 = 21$ is the standard ${{21+5-1} \choose {21}}$.
But if you want to restrict $x_i \ge 2$ the answer is ${{11+5-1} \choose {11}}$.
Note the change to 11. The mind trick here is to think of putting two ones into each variable.
That uses ten ones leaving eleven ones to distribute.

The first one is really ambiguous. Not sure about piles. Can you have an empty pile?
Are the piles distinguishable other that by number contained?

3. Originally Posted by Moo
To complete on that... If you want $x_i \ge 2$, substitute : $y_i=x_i-2 \quad (\implies x_i=y_i+2)$

Thus $y_i \ge 0$ and $21=x_1+x_2+x_3+x_4+x_5=y_1+y_2+y_3+y_4+y_5+5 \cdot 2 \implies y_1+y_2+y_3+y_4+y_5=11$

Hence the formula Plato gave =)

4. Originally Posted by Plato
The number of non-negative integer solutions to $x_1 + x_2 + x_3 + x_4 + x_5 = 21$ is the standard ${{21+5-1} \choose {21}}$.
But if you want to restrict $x_i \ge 2$ the answer is ${{11+5-1} \choose {11}}$.
Note the change to 11. The mind trick here is to think of putting two ones into each variable.
That uses ten ones leaving eleven ones to distribute.
So I'm trying to rationalize this here, and I'm having some difficulty.

In the first situation, I evaluate:

(21+5+1)! / 21!4! = 12650

This seems plausible, but it seems to me that:

4+4+8+2+3 = 21

is counted twice, once for the arrangement you see there and once for:

4+4+8+3+2 = 21

I tried dividing by 5!, but 12650 is not evenly divisible by 5!, which seemed odd to me at first.

But then I thought, because of the nature of what we're doing, taking a set of "a"s and putting "b"s in-place to separate groups, the two different "orderings" of the identical "4"s isn't double-counted, but "3" and "2" in reversed order is counted twice.

Help?

5. Originally Posted by InfinitePartsInHarmony
This seems plausible, but it seems to me that: 4+4+8+2+3 = 21
is counted twice, once for the arrangement you see there and once for:
4+4+8+3+2 = 21
Those two solutions are different if $x_4 \ne x_5$.
What is the exact wording of the question?