25 April 2013

New Puzzle!!

Flying back from St. Louis. Thought of a puzzle that incorporates my love for running in it..

There are 4 runners in a race. How many possible outcomes can be there in that race. Be aware that there can be ties. Which means two or three or even all of them might finish the race at the same time.

To explain it further, if A B C and D are the runners, one outcome could be D came first, A and C came joint second and B came last. This would be different from D coming first , A coming second, C coming third and B coming last. Got it?

For people who is reading this on FB and not on my blog, always send me your answers by personal message. I just made this up – so I do not know the answer myself, but I have an hour of flight left to figure it out 🙂

Posted April 25, 2013 by Rajib Roy in category "Puzzles


  1. By Rajib Roy on

    Kuntal, you are probably going to crack this in a nanosecond. I sent you my final number in a PM to you. Can you see if you came to the same answer?

  2. By Ashish Misra on

    4! for discrete positions 1 through 4; plus (2*3)*4C2 for 2 runners tied (4C2) in one of three possible positions (1st through 3rd, hence times 3) and other 2 runners arranging in two possible ways or the other two positions (hence times 2); plus 2*4C3 for three runners tied times 2 depending on if they are tied 1st or tied 2nd; plus 1 for a 4-way tie. Not sure what it adds up to. Have I missed anything?

  3. By Rajib Roy on

    Amy, your message puts me in a quandry. I normally ask people send the answers by PM so that others get a chance to solve it themselves. But looks like you like reviewing other people’s work. Hmmmm. I wonder if it is possible to publish certain messages later as comments…

  4. By Rajib Roy on

    Ritesh and Antara got the answers. Obviously I am assuming what Kuntal and I came up with is the right answer. I am not too confident of my answer – but I am very confident of Kuntal.

  5. By Ashish Misra on

    Whoops, schoolboy error – lamentable lack of form on my part this evening, clearly. Thanks for the endorsement though Amy, and may I second that proposal to allow for peer reviews

  6. By Baishali Chakraborty on

    Rajib Roy, Rajibda, Pl,don’t take it personally. Sudhu bangla medium rai careless hoi na, english medium er onek stupid er examples acche amar katche.may be our english are terrible, but in many cases our common sense, intelligence r higher than any medium.

  7. By Ashish Misra on

    Struggling to see that … clearly missed something. Look forward to the solution when you make it public. Cheers!

  8. By Rajib Roy on

    Here is the answer as some have figured out already.
    There are 5 possible cases – (a) each come at different times (b) two come at the same time but the other two have two different times (c) two come at the same time and the other two come together also (at a different time, of course) (d) three come at the same time and (e) four come at the same time.
    (a) This will be how many ways can four different persons arrive: 4! = 24
    (b) In this case, if you think of the two tied person as one person, there are 3! = 6 ways you can permute them. And now you can pick two people from the four people to form the tie in 4C2 = 6 ways. So there are 6×6 = 36 ways.
    (c) In this case, there is a problem of possibly double counting. The best way to think is in how many ways can I pick the two guys who came first. When I do that it is automatically decided who they other two are and what position they came in. So, it is 4C2 = 6. [If you try the same method as (b), you have to divide the final result by 2 because “picking A and B to come in first set is the same as picking “C and D” in the second set]
    (d) Think again of the three people tying as one person – there are 2! = 2 ways of permutation. And you can pick 3 out of 4 people to be in the tied set in 4C3 = 4 ways. So total is 2 x4 = 8.
    (e) Finally there is only one way they can be all tied = 1

    Total = 24 + 36 + 6 + 8 + 1 = 75 ways.


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