If the XAT centre of 4 students can be any one of the 7 cities, then calculate the probability that all the 4 students get any one of exactly 2 centres.
Total number of ways in which centres can be allocated = 74. Two centres can be chosen in 7C2 ways. The number of ways they can accommodate 4 students = 24
But this includes the two cases where one centre is having all the four students. So, no. of ways = 24 – 2 = 14. So, total number of favourable ways = 7C2 × 14
So, probability = 7C2 × =.
Hence option (b)