There are 5 students in the math club, and 3 students will be selected to compete in the math contest. how many different combinations of students are possible?

Students a b c d e
Possible ways:
ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE
Hence 10 possible combinations

Best answer that i found on yahoo ok   set :   x= length of 2nd piece in cm  x- 3= length of 1st piece in cm  x-4= length of 3rd piece in cm  then since you know they all came from the 20 cm ribbon you add them together and set them equal to 20x+ x-3 +x-4 = 20  3x - 7= 20  3x= 27  x= 9  the shortest piece is the third piece which is x- 4  x-4  9-4 = 5  the length of the smallest piece is 5 cm.  the other lengths are  9 cm  6 cm