In how many ways can a teacher arrange five students in the front row of a classroom with a total of 40 students

In how many ways can a teacher arrange 5 students in the front row of a classroom with a total of 40? 

Since order matters this ia a permutation 

40P5 

I am going to finish the problem (my way) and then show you my fast method. 
Pardon me if you already know it. 

40P5=(40)(39)(38)(37)(36)=78960960 

n= total number of objects you are choosing from 
r= the number of objects you are choosing 
n! 
nCr= 
(n-r)!*r! 
BUT I dont use itto slow 
Start with how I do a Permutation 
10P2=start with 10 and count down 2 numbers 
10P2=10*92 numbers because of the 2 after the P 
10P2=10*9=90 
8P3=8*7*6=336 
Now the Combination starts off the SAME way 
10C2=10*9 EXCEPT now you have a denominator 
to get the denominator you start with 1 and count up to 2 
to whatever the number is following the C 
10*9 
10C2==this how you set it up. Now either cancel or multiply 
1*2 
10*9 
10C2==90/2 =45 
1*2 
8*7*6 
8C3==cancel the 2*3 in the den. with the 6 in the num,=8*7=56 
1*2*3 
Special NoteOnly on combinations (not permutations) may you do the following 
10*9*8*7*6*5*4 
10C7=NO, do not continue 
1*2*3*4*5*6*7 
Subtract 10 minus 7answer is 3 
Replace the 7 with a 3 
10*9*8 
10C3==10*3*4..(i reduced/canceled)..=120 
1*2*3

It can be 8 columns because 8 times 5 =40

0=6x+2y+10 is equivalent to the simplified form: y= -3x - 5• parallel lines always have the same slope, different y intercept knowing this, we can say the slope, or m of the line parallel to y= -3x - 5 is -3.m= -3 hope this could !