## Problem Of The Week 28

**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?**

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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

milindSeptember 26, 2008 at 6:12 am

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

milindSeptember 26, 2008 at 8:54 am

57

VarunOctober 1, 2008 at 1:43 pm

whats the correct answer outtimed?

CATastrophe08October 1, 2008 at 2:49 pm

11/46

so sum is 57

outtimedOctober 1, 2008 at 3:00 pm

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

milindOctober 25, 2008 at 3:53 pm