We must find all possible groups of stones that can be selected so that the group sum is 11 .

To do this, we first consider groups of size two.

There are 5 groups of size two whose numbers sum to 11 .

These groups are: $\{1,10\},\{2,9\},\{3,8\},\{4,7\}$, and $\{5,6\}$.

Next, we consider groups of size three.

There are 5 groups of size three whose numbers sum to 11.

These are: $\{1,2,8\},\{1,3,7\},\{1,4,6\},\{2,3,6\}$, and $\{2,4,5\}$.

Can you verify that there are no other groups of size three?

Although a group of size four is possible, $(\{1,2,3,5\})$, it is not possible to form two more groups whose numbers sum to 11 using the remaining stones $(\{4,6,7,8,9,10\})$.

Therefore, only groups of size two and size three need to be considered.

Next, we must count the number of ways that we can select three of the ten groups listed above such that no number is repeated between any of the three groups.

We may consider four cases or ways that these 3 groups can be chosen.

They are: all 3 groups are of size two, 2 groups are of size two and 1 group is of size three, 1 group is of size two and 2 groups are of size three, and finally all 3 groups are of size three.

Case 1: all 3 groups are of size two

Since there is no repetition of numbers between any of the 5 groups having size two, choosing any 3 groups will produce a solution.

There are 10 possible solutions using groups of size two only.

These are:

$$
\begin{array}{ll}
\{1,10\},\{2,9\},\{3,8\} & \{1,10\},\{4,7\},\{5,6\} \\
\{1,10\},\{2,9\},\{4,7\} & \{2,9\},\{3,8\},\{4,7\} \\
\{1,10\},\{2,9\},\{5,6\} & \{2,9\},\{3,8\},\{5,6\} \\
\{1,10\},\{3,8\},\{4,7\} & \{2,9\},\{4,7\},\{5,6\} \\
\{1,10\},\{3,8\},\{5,6\} & \{3,8\},\{4,7\},\{5,6\}
\end{array}
$$

Case 2: 2 groups are of size two and 1 group is of size three

Since there is repetition of numbers between some of the groups of size two and some of the groups of size three, we need to take care when making our choices.

To systematically work through all possible combinations, we will first choose a group of size three and then consider all pairings of groups of size two such that there is no repetition of numbers in either group of size two with the numbers in the group of size three.

This work is summarized in the table below.

| 1 group of size three | 2 groups of size two |
| :---: | :---: |
| $\{1,2,8\}$ | $\{4,7\},\{5,6\}$ |
| $\{1,3,7\}$ | $\{2,9\},\{5,6\}$ |
| $\{1,4,6\}$ | $\{2,9\},\{3,8\}$ |
| $\{2,3,6\}$ | $\{1,10\},\{4,7\}$ |
| $\{2,4,5\}$ | $\{1,10\},\{3,8\}$ |

There are 5 possible solutions using 2 groups of size two and 1 group of size three.

Case 3: 1 group is of size two and 2 groups are of size three

There is considerable repetition of numbers between the 5 groups of size three, $\{1,2,8\},\{1,3,7\}$, $\{1,4,6\},\{2,3,6\}$, and $\{2,4,5\}$.

In fact, there are only two groups whose numbers have no repetition.

These are $\{1,3,7\}$ and $\{2,4,5\}$. Can you verify that this is the only possibility?

At this point, we are unable to choose a group of size two such there is no repetition with the groups $\{1,3,7\}$ and $\{2,4,5\}$.

Therefore, there are no solutions using 1 group of size two and 2 groups of size three.

Case 4: all 3 groups are of size three

In Case 3, we found that there were only 2 groups of size three that had no repetition of numbers.

Therefore it is not possible to find 3 groups of size three without repeating numbers.

That is, Case 4 produces no solutions.

Thus, the total number of possible arrangements of the stones into three groups, each having a sum of 11 , is $10+5=15$.

ANSWER: (E)
Stones are numbered $1,2,3,4,5,6,7,8,9,10$. Three groups of stones can be selected so that the sum of each group is 11 . For example, one arrangement is $\{1,10\},\{2,3,6\}$, $\{4,7\}$. Including the example, how many arrangements are possible?
(A) 13
(B) 16
(C) 11
(D) 12
(E) 15