Subject: Mathematics — Permutations and Combinations
Question: Let $S = \{ 1,2,3,\ldots,10\}$. Consider the set\
$X = \{ R:R$ is an equivalence relation on the set $S$ such that $R$ has
exactly 42 elements $\}$.\
Then the number of elements in X is $\_\_\_\_$ .
Answer: . 2520.00\
Solution: The total number of such relations is 2520.
Here is how you can think about it step-by-step:\
An equivalence relation is just a way of breaking a group into
smaller, separate teams (called equivalence classes). Within any team,
every member connects to everyone else, including themselves.\
This means a team with \'n\' people creates $n^{2}$ total
connections.\
For our problem, we have 10 people in total. We need to split them
into teams so that the sum of the squared team sizes equals exactly
42.
We need to find combinations of numbers that add up to 10, but whose
squares add up to 42. If you test out different numbers, only two
combinations actually work:\
Option A: One team of b, one team of 2 and two teams of 1. Check total
people: $(6 + 2 + 1 + 1 = 10)$ check total connections:
$\left( 6^{2} + 2^{2} + 1^{2} + 1^{2} = 36 + 4 + 1 + 1 = 42 \right)$.\
Option B: One team of 5, one team of 4 and one team of 1. Check total
people : $(5 + 4 + 1 = 10)$ Check total connections :
$\left( 5^{2} + 4^{2} + 1^{2} = 25 + 16 + 1 = 42 \right)$.\
Now we just count how many ways we can sort our 10 people into these
two setups.\
Fo option $A(6,2,1,1)$ : First, pick 6 people out of 10 for the big
team.\
Then, pick 2 out of the remaining 4 for the second team. The last 2
people automatically form their own single-peson teams.\
Because the two single person teams are identical in size, we divide
by 2 to avoid double-counting them.\
Ways $= \frac{10!}{6!2!1!1!} \times \frac{1}{2!} = 1260$\
For option B\
Pick 5 people out of 10 fo the first team, then 4 out of the remaining
5 for the second team.\
The last person is left by themselves.\
Ways $= \frac{10!}{5!4!1!} = 1260$\
Total count\
Add the tow possibilities together : $(1260 + 1260 = 2520)$