GATE CSE
First time here? Checkout the FAQ!
x
+3 votes
114 views

How many different equivalence relations with exactly three equivalence classes are there on a set with 5 elements

  1. 10
  2. 15
  3. 25
  4. 30

 

asked in Set Theory & Algebra by Veteran (37.9k points)  
edited by | 114 views
25 is the answer ??
yes explain plz

1 Answer

+8 votes
Best answer
  • One Euivelence relation creates one partition.
  • S = $\left \{ a,b,c,d,e \right \}$ ,here one possible parition is = $\left \{ \left \{ a,b \right \} , \left \{ c,d \right \} , \left \{ e \right \} \right \}$ . Tis partition has 3 groups. Corresponding EQ relation has 3 EQ classes. Note that this partition is unordered..
  • So, no of different unordered partitions = No of equivalence relations.

Here the condition is we need only 3 equivalence classes. So, the partition has to be done in 3 unordered groups.

 

$$\begin{align*} \frac{5!}{1!1!3!}.\frac{1}{2!} + \frac{5!}{2!1!2!}.\frac{1}{2!} = 10 + 15 = 25 \\ \end{align*}$$ 

In case if we need maximum possible EQ relations on 5 elements.

$\begin{align*} 5 &\rightarrow 5\Rightarrow \frac{5!}{5!} = 1\\ \\ \hline \\ &\rightarrow 4 \qquad 1 \Rightarrow \frac{5!}{4!1!} = 5 \\ \\ \hline \\ &\rightarrow 3 \qquad 2 \Rightarrow \frac{5!}{3!2!} = 10 \\ \\ \hline \\ &\rightarrow 3 \qquad 1 \qquad 1 \Rightarrow \frac{5!}{3!1!1!}*\frac{1}{2!} = 10 \\ \\ \hline \\ &\rightarrow 2 \qquad 2 \qquad 1 \Rightarrow \frac{5!}{2!2!1!}*\frac{1}{2!} = 15 \\ \\ \hline \\ &\rightarrow 2 \qquad 1 \qquad 1 \qquad 1 \Rightarrow \frac{5!}{2!1!1!1!}*\frac{1}{3!} = 10 \\ \\ \hline \\ &\rightarrow 1 \qquad 1 \qquad 1 \qquad 1 \qquad 1 \Rightarrow \frac{5!}{1!1!1!1!1!}*\frac{1}{5!} = 1 \\ \\ \hline \\ &\Rightarrow \text{Total} = 1+5+10+10+15+10+1 = 52 \end{align*}$

The procedure for partitioning is , indistinguishable $\rightarrow$ indistinguishable but because of the distinct elements we need to further evaluation.

Graphically we were interested in the rectangle shown below containing 25 EQ relation or partitions.

answered by Veteran (36.8k points)  
edited by
This is also Bell Number for maximum case, right ?
yes :) $B_5$
thanks for ur detailed ans  , but why u have divided it by 2! in case of 3 1  1   and 2  2  1   

and by 3!  in  2   1   1   1 and so on
if we have a set {a,b,c,d} and two indistinct boxes

To allocate {a,b,c,d}...two-and-two,  in those two boxes we have only 3 choices. $\frac{4!}{2!2!}*\frac{1}{2!} = 3$

those are $\left \{ \left \{ a,b \right \} ,\left \{ c,d \right \} \right \} , \ \left \{ \left \{ a,c \right \} ,\left \{ b,d \right \} \right \} , \ \left \{ \left \{ a,d \right \} ,\left \{ b,c \right \} \right \}$

No need to put a in box2 and count again. because boxes are indistinct.

In analogy to these boxes, those groups of a partition are also same,only the elements inside a group are distinct.
Top Users Jan 2017
  1. Debashish Deka

    8968 Points

  2. sudsho

    5326 Points

  3. Habibkhan

    4798 Points

  4. Bikram

    4532 Points

  5. Vijay Thakur

    4486 Points

  6. saurabh rai

    4222 Points

  7. Arjun

    4196 Points

  8. santhoshdevulapally

    3808 Points

  9. Sushant Gokhale

    3596 Points

  10. Kapil

    3486 Points

Monthly Topper: Rs. 500 gift card

19,212 questions
24,104 answers
53,150 comments
20,319 users