I am not sure if I am right.. And I have given the answer in numerical terms..

I would be happy if you can confirm if atleast the approach is right..

My line of thinking goes like this..

For the first question, (before reading the second question), we are lead to believe that the rings are not identical. But after reading the second question, if we assume that the all the rings in the first question are identical, then I think there is just one way of wearing 5 rings in the four particular fingers of right hand.

So the ans is 1

(since the fingers are particular we dont get combinations there, and since the rings are identical we dont get combinations there also)

For the second question, given the rings are different and we have 4 particular fingers, it goes like

for the 1st ring u ve got 4 free fingers

for the 2nd ring u ve got 3

for the 3rd ring u ve got 2

for the 4th ring u ve got 1

4X3X2X1 = 24 possiblities. BUT, there is one ring left out. This means that for ever ring left out there aer 24 possibilties.

In other words If R1 is selected first there are 24 possibilities, if R2 is selected first there are 24 possibiltes and so on..

So ans is 24x5 = 120For the third question we are free to choose any 4 of the 5 fingers.. So, it just the matter of multiplying the ans for the second question with the no. of finger combinations.

No of possible finger combinations are 5! / (4!x1!) = 5

So the ans is 120x5 = 600ny comments guys ??

Bunuel wrote:

No one else wants to try? I don't have OA for these questions (well I have for the first one but as I said I disagree) so can only post my answers if no more attempts.