Problem Analysis #15

[THOUGHT] To solve this problem, we need to understand the properties of reflexive and symmetric relations.

A relation R on a set is said to be reflexive if for every element x in the set, (x, x) ∈ R. It means that every element is related to itself.

A relation R on a set is said to be symmetric if (a, b) ∈ R then (b, a) ∈ R for all elements a and b of the set. It means if a is related to b, then b is also related to a.

Given that our set {a, b, c, d, e, f} has 6 elements, we need to find how many relations satisfy these two conditions and have exactly 10 pairs.

1. Reflexive property: Since (a, a), (b, b), ... , (f, f) must be in R for it to be reflexive, there are 6 "guaranteed" pairs.
2. Symmetric property: For each pair (x, y) that is in R, the symmetric pair (y, x) will also be in R because of symmetry.

Now we have a total of 6 + 10 - 1 = 15 "slots" to fill:

• The first 6 are for the reflexive pairs.
• The next 10 are for the additional symmetric pairs that make up our relation (since each pair contributes 2 elements).
• But, because (x, x) is both a reflexive and a symmetric pair, we must subtract one.

The question now becomes: How many ways can we choose 10 distinct "slots" from these 15?

This is a combination problem. The number of combinations C(n, r) = n! / [r!(n - r)!] gives the number of ways to select r items from a set of n without regard to order.

Here, n = 15 (total slots), and we want to choose r = 10 (for our relation's additional pairs).

So, C(15, 10) = 15! / [10!(15 - 10)!] = 15! / (10! * 5!) = 3003.

Therefore, there are 3003 different relations R on the set {a, b, c, d, e, f} that are both reflexive and symmetric, with exactly 10 elements.

\boxed{3003}