Don’t you think there will be 4 factorial ways of arranging the first 4 shoes?
Whoops, yes there should, thanks for pointing out 🙂 Edited…
Sir, Need an advice. How to proceed for the Binary logic questions like truth-teller,liar and alternators case.
E.g. In a family, there are 4 members Bingo, Tingo, Mingo and Pingo. Each member of that family belongs to one of truth-tellers, liars & the alternators. They made the following statements when they are asked about their profession. And no 2 persons are of the same profession.
Bingo: I am a truth-teller, I am the manager, Tingo is the CA.
Mingo: I am a truth-teller, I am the Engineer, Professor is an alternator.
Tingo: I am a truth-teller, I am the professor, Pingo is the Engineer.
Pingo: I am an alternator, I am the professor, Bingo is a liar.
Do we need to go with hit & trial cases only or we can somehow cut short the way to minimize the time while solving.
Pretty much trial and error. Having said which, such questions haven’t appeared in CAT for ages. So don’t waste time on them for now if you are not already comfortable with them!
Could you please solve this question:
Number of arrangement of eight shirts in a row ( there are four different styles of shirt, two identical ones of each particular style) such that no two identical shirts are next to one another.
Shouldn’t the logic be same as the above problem on 4 pairs of shoes?
Similar but not same, as the two shirts of a particular pair are identical but in the question in my post, the two shoes of a pair are distinct (left and right).
should it be 4! *5c2 * 7c2 ?
I think a few arrangements have been left out.
Eg: The arrangement AbaBCdcD is valid but cannot be made by the method described.
I have a doubt here in 1st ques, why are we go on adding the new places in each step..? Like ‘a’ could have been placed between C and D, why did you make two extra vacant places besides ‘a’ after placing it…?
Bro if the total no of arrangements are 8!/2!^4 i.e., 2520, then how come with restriction of same pair never come together is greater than the total no of arrangements as 3*4*5*6*4! is 8640
Total number of arrangements is 8!. It would be 8!/(2!)^4 only if the two shoes of one type are identical. But they aren’t. However, as pointed out by another commenter above, there are a few cases left out, so even my solution is incomplete. If I can think of a better approach, will update it!
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