The number of possible combinations is given by
... C(18, 3) = 18!/(3!(18-3)!) = 18·17·16/(3·2·1) = 816 . . . . possible combinations
_____
There are 18 ways to choose the first one; 17 ways to choose the second one, and 16 ways to choose the 3rd one. The same 3 students can be chosen in any of 3! = 6 different orders, so the product 18·17·16 must be divided by 6 to get the number of possible combinations in which order doesn't matter.
But i cant see the picturess
Any
in this set will be real numbers that are both less than
and greater than
. But that's not possible, so this set is empty.
Answer:
30
Step-by-step explanation:
3x - 27 = x +11
3x -27 +27 = 11 + 27
3x - x = x -x + 38
2x = 38
2x ÷ 2 = 38
X = 19
19 × 3 -27 = 30
