# DM-Sets,Relations and Functions-Q11

+1 vote

Let $A = \{1,2,3,4,5,6,7\}$ The number of symmetric relations on set A which contains exactly 4 elements?

reshown Jul 6

a=[1,2,3,4,5,6,7]

a*a={(1,1),(2,2),(3,3).................................(7,7),  (1,2)(2,1)  (1,3)(3,1)...............................................................}

no of diagonal elements=7

no of non diagonal elements=42

now since it is symmetric relation we will take all the elements like (a,b) and (b,a) into pairs. so no of pairs in above question will be =42/2=21

now remember each pair contain two elements

now in symmetric relation diagonal elements can be there in any order or any number of times.

ATQ we have to form the relations with elements 4

so let take the cases:

case 1: there are no diagonal elements and 2 pairs     7c0 * 21c2= 210

case 2: there are  two diagonal elements and 1 pair    7c2 * 21c1=441

case 3: there are 4 diagonal elements and 0 pair    7c4* 21c0 = 35

total=210+441+35=686
answered Jul 6 by (4,600 points)
selected Jul 8
suggest me if i m wrong
perfect solution
i thought i was wrong when someone gave me down vote.
686s the correct one.
answered Jul 6 by (18,160 points)
edited Jul 6

686 is correct one.

answered Jul 6 by (18,160 points)
good problem
Total 7*7=49 pair
In which 7 is like (X,X) type and 42 is (x,y) (y,x) like that if we select X,y than y,X automatically selected
So in my hand ony 21 selection 21 automatically depend upon my selection like 1,3 we select than 3,1 automatically selected I think now clear.
i also did this way, best question of this test
Yes realy v good problem
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