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FoolForMath
Fool's problem of the day, A panel of eight umpires (A, B, C, D, E, F, G and H) is selected for officiating all the Cricket matches played in 2002. The countries which they represent (belong to) are P, Q, R, S, T, U and V, not necessarily in the same order. (i) In the panel, B and D only belong to the same country i.e., T, while the others represent different countries. (ii) A belongs to one of the subcontinent countries i.e., Q or U, while G belongs to the other. (iii) The match between R and V is officiated by E and H. (iv) H is does not belong to country P. Note:- The umpires of the same country as the 2 countries that are playing the match cannot officiate in the match. \(Q_1\): If a match is played between countries Q and V, then how many different combinations of umpires for that match are possible? \(Q_2\): If in the above question, it is given that E and H are good friends, and always officiate together in a match, then how many different combinations of umpires are possible for the match between Q and V? Attribution: I faced this question in a selectional test conducted by my coaching institute IMS ( http://www.imsindia.com/ )
Dont know if I'm way off here, but I think its Q1;(9) and Q2;(0)
Damn, will try again
Is it just me, or is the majority of this information extraneous?
Yea nbouscal that is what i feel too. Thats why i don't think i got the right answer My answer: Q1. 49 Q2. 25
I was getting Q1. 15 and Q2. 7
May we know the real answer. Thanks!
The only suitable umpires I could be sure of are B, D, E and H giving the answers: Q1. 6 combinations Q2. 2 combinations
I am getting Q1: 15, Q2: 7 as well. I had to draw an image to visualize how it worked, which is attached. lol
why you call it a FOOL's problem I think it should be Einstein's problem
Ya @Glen_McGrath
I'm getting Q1 = 24. Q2 = 7. May be wayy off.