Let the set of all relations R on the set {a, b, c, d, e, f}, such that R is reflexive and symmetric, and R contains exactly 10 elements, be denoted by 𝒮. Then the number of elements in 𝒮 is __________________.

The correct answer is 105.

Let us break down the solution step-by-step:

Step 1: Understand the structure of the relation R
The relation R is defined on the set A = {a, b, c, d, e, f}, which contains 6 elements, so:
$n\left(A\right)=6$
A relation R on A is a subset of the Cartesian product A × A.

Step 2: Apply the reflexivity condition
For R to be reflexive, it must contain the element (x, x) for every x in A. This means the following 6 ordered pairs must belong to R:
{(a, a), (b, b), (c, c), (d, d), (e, e), (f, f)}
These account for exactly 6 elements of the relation R.

Step 3: Apply the symmetry condition and element count
The problem states that R contains exactly 10 elements. Since 6 of these elements are the reflexive pairs, we need to choose:
$10-6=4$
additional elements from the off-diagonal pairs (where x ≠ y).
Since R is symmetric, if any ordered pair (x, y) is in R, then its counterpart (y, x) must also be in R. Since x ≠ y, these elements always occur in pairs of the form {(x, y), (y, x)}.
Since we need exactly 4 off-diagonal elements, we must select exactly:
$\frac{4}{2}=2$
symmetric pairs of the form {(x, y), (y, x)}.

Step 4: Calculate the number of available symmetric pairs
The number of distinct, unordered pairs of distinct elements we can choose from the 6-element set A is given by:
$\left(\genfrac{}{}{0ex}{}{6}{2}\right)=\frac{6\times 5}{2}=15$
Thus, there are 15 possible symmetric pairs of the form {(x, y), (y, x)} available to choose from.

Step 5: Calculate the number of ways to choose the pairs
To construct a relation R in the set 𝒮, we must choose exactly 2 symmetric pairs out of the 15 available. The number of ways to do this is:
$\left(\genfrac{}{}{0ex}{}{15}{2}\right)=\frac{15\times 14}{2}=105$
Therefore, the number of elements in the set 𝒮 is 105.