Question

48 students have to be seated such that each row has the same number of students as the others. If at least 3 students are to be seated per row and at least 2 rows have to be there, how many arrangements are possible?

Answer 1

Answer:

Step-by-step explanation:

48 = 2^4 * 3^1

now separating the factors:

x*y = 1*48, 2*24, 3*16, 4*12, 6*8, 8*6, 12*4, 16*3, 24*2, 48*1.

let x=rows and y=no. of students, therefore 10 combinations are possible

according to the question x>=2 and y>=3, therefore 7 combinations is the ans.

Hope it helps you...