-- He must have at least one of each color in the case, so the first 3 of the 5 marbles in the case are blue-green-black.
Now the rest of the collection consists of
4 blue 4 green 2 black
and there's space for 2 more marbles in the case.
So the question really asks: "In how many ways can 2 marbles be selected from 4 blue ones, 4 green ones, and 2 black ones ?"
-- Well, there are 10 marbles all together. So the first one chosen can be any one of the 10, and for each of those, the second one can be any one of the remaining 9 .
Total number of ways to pick 2 out of the 10 = (10 x 9) = 90 ways.
-- BUT ... there are not nearly that many different combinations to wind up with in the case.
The first of the two picks can be any one of the 3 colors, and for each of those, the second pick can also be any one of the 3 colors.
So there are actually only 9 distinguishable ways (ways that you can tell apart) to pick the last two marbles.
You first identify what supplementary angles are,angles that add up to 180. 5x+2+8x+9=180 17x+11=180-11 17x=169 x=169divided by 17 x=9.9411 x=10 when rounded into the next value 8x*10=80+989
A=P×(0.8) A is the price after the markdown P is the original price 0.8 is the multiplier because 1-(80/100)=0.8 80/100 is the same as 80%, which is the remaining value when 20% of the value of the item is deducted.