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Problem Of The Week 28

with 6 comments


Twenty five of King Arthur’s knights are seated at their customary round table. Three of them are chosen – all choices being equally likely – and are sent of to slay a troublesome dragon. Let p be the probability that at least two of the three had been sitting next to each other. If p is written as a fraction in lowest terms, what is the sum of the numerator and the denominator?

Written by Implex

September 25, 2008 at 8:51 am

6 Responses

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  1. no. of ways of selecting 3 out of 25=25C3
    Now, taking the adjacent pair as 1/2 , third can be any1 from 3 to 25. So, 23 ways
    for 2/3, third can be any1 from 4 to 25=22 ways.
    So, similarly up to 23/24 and third can be 25.
    So, no. of favourable combinations are 1+2+….23
    Prob= 23*24/(2*25C3)=3/25
    So, ans is 28

    milind

    September 26, 2008 at 6:12 am

  2. I made a mistake here.
    Didn’t count few numbers.
    For 2 and 3, third can take 22 values
    For 3/4, third number can take 1 and values from 5 to 25=22
    So, in all 23*22+23=23^2
    So, prob=23*23/(25C3)=23*23/(23*24*25/3)=23/100
    Ans is 123

    milind

    September 26, 2008 at 8:54 am

  3. 57

    Varun

    October 1, 2008 at 1:43 pm

  4. whats the correct answer outtimed?

    CATastrophe08

    October 1, 2008 at 2:49 pm

  5. 11/46
    so sum is 57

    outtimed

    October 1, 2008 at 3:00 pm

  6. Did calculation mistakes..
    Let’s revisit..
    For pair 1/2 third no. can be from 3 to 25. Total=23
    pair 2/3 third no. can be from 4 to 25. Total=22
    pair 3/4 third no. can be from 5 to 26 and 1. Total =22
    Similarly for all pairs upto 24/25.
    And, for pair 25/1 third can be 3… 23. Total=21
    So, total cases =23+ 22*23+ 21=550
    And total possible selections =25C3.
    So prob=550/50*46=11/46. So, 57

    milind

    October 25, 2008 at 3:53 pm


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