Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
2300 you mover the decimal over 3 spaces
The answer is b cause none of x is the same number
Answer:
x^2 - 11x + 24
Step-by-step explanation:
Foil the equation
x^2 - 3x - 8x + 24
x^2 - 11x + 24
Answer:
Multiply the first two binomials: (x+2) & (x+4) using the FOIL method (First, Outer, Inner & Last).
(x+2)(x+4) = x^² + 4x + 2x + 8 = x^² + 6x + 8
Then we multiply the trinomial that we got from the first stage and multiply that by the remaining binomial:
(x^² + 6x + 8) (x + 4)
= x^³ + 4x^² + 6x^² + 24x + 8x + 32
Collect the like terms together to give you:
x^³ + 10x^² + 32x + 32
