There are 18 students participating in a spelling bee. how many ways can the students who go first, second, third, and fourth be chosen?

a.
73,440 ways
c.
4,896 ways
b.
1,028,160 ways
d.
13,366,080 ways

a. 73,440 ways

Step-by-step explanation:

To find the number of ways to choose first, second, third and fourth in a spelling bees given that 18 student participate in the spelling bee, we will simply use permutation.

Thus, the standard formula is; n  =  n !   /  (n-r)!

From the question above, our n is equal to eighteen and  our r is equal to four That is; n= 18 and r = 4

Inserting these value into the formula

n  =  n !   /  (n-r)!

18    =    18 !  /   (18 - 4) !

             = 18 !   /   14 !

             = 18 × 17 × 16 × 15 × 14!      /     14!

The 14! on the numerator will cancel-out the 14! on the denominator, hence;

18  =  18 × 17×16×15

           =73,440

Therefore  we can choose the students who goes first, second, third and fourth position in 73, 440 ways.

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.