If the examination center of 4 students can be in any one of the 7

4 students, 7 cities, students to placed in any one of the exactly two chosen cities.

1. Total number of ways :- basically any student can be placed in any of the cities. therefore,7*7*7*7 no of ways.

2. possible no of ways :- first we select only two cities, which can be chosen in 7C2 =21 ways.

now,
Case 1: 3 students in the first city, 1 student in the other city
4C3 * 1C1 = 4.
Case 2: 2 students in each city
4C2 * 2C2 = 6.
Case 3: 1 student in the first city, 3 students in the other city
4C1 * 3C3 = 4.
Total ways to distribute the 4 students = 4+6+4 = 14.
Thus, there are 21 pairs of cities that could be chosen, and within the chosen pair, 14 ways to distribute the 4 students.
Total options for the 4 students = 21*14.

required probability = 21*14/ 7*7*7*7 = 6/49

answer B

Letting A, B, C, D, E, F and G denote the cities, we are looking for the probability that a selection of four cities (such as A-B-C-D or D-E-F-G) includes exactly two different cities.

Since there are 7 cities, the number of different selections is 7^4 (repetitions allowed).

Let’s first find the number of ways where the selection only includes A and B. Since there are two choices for each student, there are 2 x 2 x 2 x 2 = 16 selections which include only A and B. However, one of these selections is A-A-A-A and one is B-B-B-B, neither of which is possible. Therefore, there are 16 - 2 = 14 selections which include only A and B and at least one of each.

Next, we note that the requirement “all students get any one of exactly 2 centers” can also be satisfied through selections such as C-D-C-D or E-F-F-E. Thus, we need to determine the number of ways we can select two letters out of seven. There are 7C2 = (7 x 6)/2 = 21 ways to make such a selection.

Since there are 7^4 possibilities and 14 x 21 of them are favorable, the required probability is:

(14 x 21)/7^4 = (2 x 3)/7^2 = 6/49

Answer: B

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