Computing the value of P

Question

Answer 1

 From i=1 to n

 "if(j<i)" this condition true till only for every j=i-1 times
 So, P=0+1+2....+(n-1)=n(n-1)/2

and for i=(n+1) to (2n-1)

"if(j<i)" This will true n-1 times 
P= n(n-1)

Total value of P=n(n-1)+(n(n-1))/2=3(n²-n)/2
 

  

Comments

p=p+1 runs 0,1,2,3,3 which is close to B bt not exactly B plz chck n=6 i=1 2 3 4 5 j always runs 1 to 3 for each val of i

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All options are wrong. I'm not saying B option is correct.

Answer 2

Let's calculate number of pairs (j,i) such that j<i.

(1,2),(1,3),...,(1,2n) = Total 2n-1 pairs.

(2,3),(2,4),...,(2,2n) = Total 2n-2 pairs.

.

.

.

(n,n+1)(n,n+2),...,(n,2n) = Total 2n-n = n pairs.

Therefore Total pairs overall = n(n+2n-1)/2.

Comments

i think u calculated it for

(i<j)
isn't it ?

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I've edited. I calculated j<i only.

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but in the code for each i<2n, j is running till n
so for i=1 , j  should be vary upto n
plss correct me if i m wrong ?

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Bro, I've written (j,i) j before i. And j is UpTo n. Yes. Right. i<2n I've taken I<=2n.