For mr. fujimoto's class, there are 7 possible assignments. however, students only have to choose 5 assignments to complete and submit. if the order in which the students submit the assignments doesn't matter, how many different combinations of 5 are there?

The formula for combinations is
C=n!/[(n-r)!*r!]
where
n is the total number of objects you choose from
r is the number that you choose to arrange

in this problem
n=7 assignments

r=5 assignments

C=7!/[(7-5)!*5!]> C=7!/[2!*5!]---> (7*6*5!)/(2*5!)> C=21

the answer is

21

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!

it would not be increasing by nothing that i can see probably by 25 percent that is all i can sat\y for now if i have any thing else to say i will edit this.

step-by-step explanation: