Three students can be selected from 5 in (5 x 4 x 3) = 60 ways.BUT there are (3 x 2) = 6 ways to select the same 3 students.So there are only 60/6 = 10 different groups of 3 studentsthat can be selected from a pool of 5.
There are 3 ways. 30/56, 30 to 56, and 30 : 56.
-5
There are 30 ways of selecting the first student leaving 29 ways to select the second for a total of 30 X 29 = 870 ways
The answer is 12C5 = 12*11*10*9*8/(5*4*3*2*1) = 792
35
(5x4x3)/(3x2x1) = 10
im a webkin
-1
12C3 = 12*11*10/(3*2*1) = 220
8C3 = 8*7*6/(3*2*1) = 56
Three students can be selected from 5 in (5 x 4 x 3) = 60 ways.BUT there are (3 x 2) = 6 ways to select the same 3 students.So there are only 60/6 = 10 different groups of 3 studentsthat can be selected from a pool of 5.
2 boys from 6 = 6*5/2 = 15 ways 2 girls from 3 = 3*2/2 = 3 ways All in all, 15*3 = 45 ways.
The number of ways is 10C5 = 10!/(5!*5!) = 10*9*8*7*6/(5*4*3*2*1) = 252
560.
There are 3 ways. 30/56, 30 to 56, and 30 : 56.
-5