Michael has 7 green, 3 red and 9 blue M&Ms. If he selects 5 M&Ms:
a. In how many different ways can he select at least 2 blue M&Ms?
c(19,2)= 171
b.In how many ways can he select exactly one blue and one red M&M?
1b x 1r x 5g
c(9,1) x c(3,1) x c(7,5)= 567
c.In how many ways can he get exactly 4 of the same color?
c(7,4) x c(12,1) + c(9,4) x c(10,1)=1680
Only your last question is correct
a) at least 2 blue ----> 2 blue, or 3 blue or 4 blue or 5 blue
or
total - 0 blue - 1 blue
number with no restriction = C(19,5) = 11628
number with 0 blue = C(10,5) = 252
number with 1 blue = C(9,1) x C(10,4) = 1890
number with at least 2 blue = 11628-252-1890
= 9486
b) exactly 1 blue and 1 red (and 3 from green)
= C(9,1) x C(3,1) x C(7,3)
= 945
c) 4 of same colour ---> 4 green and 1 from red-blue OR 4 blue and 1 from green-red OR 4 from gree, but we have only 3 green, so that is not possible
number = C(7,4) x C(12,1) + C(9,4) x (10,1) + 0
= 420 + 1260
= 1680 <------ you had that.