in a five character password the first two characters must be digits and the last three characters must be letters if no characters are allowed to repeat how many unique passwords are possible

1,404,000 unique passwords are possible.

Step-by-step explanation:

The order in which the letters and the digits are is important(AB is a different password than BA), which means that the permutations formula is used to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

In this question:

2 digits from a set of 10(there are 10 possible digits, 0-9).

3 characters from a set of 26. So

1,404,000 unique passwords are possible.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

-8/2y-8=5/y+4-7y+8/y^2-16

we move all terms to the left:

-8/2y-8-(5/y+4-7y+8/y^2-16)=0

we can not solve this equation