Another approach:

1st : Select 3 students from the total of 4*2 - 8 = 8C3 = 56. (This includes 2 students from same school as well so this has to be subtracted)

2nd: Two students from the same school , so select 1 school from 4 (4C1) * from this school two students are selected so (2C2) = 4c1*2c2 = 4.
Now, remaining 1 student can be selected from remaining 6 students = 6c1 = 6
Total no of ways = 4 * 6 = 24

So the answer is 1st - 2nd = 56 - 24 = 32.

Acknowledge if you like the answer !!

Suppose there are 4 schools.

M N O P

Now each school sends 2 students as follows;

AA BB CC DD

Total of 8 students.

So the first students can be selected in 8 ways

the second students can be selected in 6 ways (because only one students can be selected from each school.So AA or BB is out)

In the same way the third student can be selected in 4 ways

Total no of selection 8*6*4

Now divide it by 3! (As the sequence can be repeated in 6 ways)

So the answer is = 8*6*4/3! =32

So, the answer is B.

\(C^3_4*2*2*2=32\), where \(C^3_4\) represents the number of ways to select which 3 schools out of the 4 will delegate the students for the debate and 2*2*2 represents the choices of the each school to delegate one of the two their students.

Answer: B.
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Since no two students can be from same b school, treat each b school as single entity. As we have to select 3 b schools,number of combination from different b school will be 4c3 ie. 4 ways (1 student from each b school , so we need 3 b schools from 4 b schools)

Now from selected 3 b schools, 1 student has to selected and we have 2 students in each b schools so we have 2 possibilities from each. ie 8 selections( 2x2x2)

so total will be 4 x8 =32

Answer is B
4c3 to select schools
Combination to select student from each school is 2
for 4 schools combination will be 8

Answer = (2*4)4c3 = 32

We are given that there are 2 students from each of the 4 different B-schools (or 8 people) and we need to determine the number of ways to choose 3 students in which no 2 people are from the same school. So this is a special combination problem. Before we tackle this problem, let’s review a combination problem with no restrictions.

With no restrictions, the number of ways to choose 3 people from 8 is 8C3, which is calculated as follows:

8C3 = 8!/[3!(8-3)!] = (8 x 7 x 6)/3! = 56

In this (special combination) problem, 3 students are chosen in which no students from the same school can serve together. The first person could be any one of the 8 students. However, once a student is selected, the other student from the same school cannot be selected. This reduces the choice of the second student to 6 possible students. Once the second student is chosen, the other student from the same school cannot be selected. This reduces the number of students who could be chosen as the third student to 4. Therefore, the number of ways of choosing these 3 students is:

(8 x 6 x 4)/3! = 32

Answer: B
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first decide the no. of slots we need to make here ,so here we need to select 3 students so there slots
88
now we have total 8 students and no 2 students should be from the same school , so selecting ways will be 8x6x4 , now divide by no. of slots that are 3 , as order does not matter here, so we get 32 as the answer

answer : B

Dear Sir ScottTargetTestPrep,
Could you please clarify why you divided 3! ?
Thank you