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I solved it by taking matrix ,I am confused in commutative

closed as a duplicate of: GATE1989-1-v

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@hemant means aij=aji 1<=I<=n and 1<=I<=n (means rows and columns will be equal because binary relation formed on same sets element ) and whatever we will put in lower half same entries will be there in upper half ,

Thnxx for making it clear :)
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But answer to this question is c)they are taking n*n*n*n*.......*n time for diagonal

=N^n so multiply this factor too
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@Prince Sindhiya

can you post their solution?

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See this

+1

+1 vote

Total no of elements = $n^2$

no of diagonal elements = n

so total no of non diagonal elements = $n^2$ - $n$

half of them will be commutative to other half=> ($n^2$ - $n$ $)/$$2$

option b?

by Loyal (6.7k points)
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You can also include diagonal elements from set S

S = { 1,2,3.....N}

1*1 = 1*1

2*2 = 2*2

..

N*N = N*N

N + (N 2 - N) /2  = N 2 + 1 / 2