14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.

Total no of ways = (14 – 1)! = 13!

Number of favorable ways = (12 – 1)! = 11!

So, required probability = $\frac{11!\times 3!}{13!}$ =  $\frac{39916800\times 6}{6227020800}$ = $\frac{24}{625}$