The answer is thanks a lot
Answer:
4032 different tickets are possible.
Step-by-step explanation:
Given : At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the first two races. If the first race runs 9 horses and the second runs 8.
To find : How many different tickets are possible ?
Solution :
In the first race there are 9 ways to pick the winner for first and second place.
Number of ways for first place - 
Number of ways for second place - 
In the second race there are 8 ways to pick the winner for first and second place.
Number of ways for first place - 
Number of ways for second place - 
Total number of different tickets are possible is


Therefore, 4032 different tickets are possible.
Answer:
The answer is explained below
Step-by-step explanation:
A. How many ways can you distribute 4 different balls among 4 different boxes
The number of ways in which 4 different balls can be distributed among 4 different boxes is 
B. How many ways can you distribute 4 identical balls among 4 identical boxes?
The number of ways in which 4 identical balls can be distributed among 4 identical boxes = P(4,1) + P(4,2) +P(4,3) + P(4,4) = 1 + 2 + 1 + 1 = 5 ways
Where P(k,n) is the number of partitions that k can be divided into n parts
P(4,1) = 4 = 1
P(4,2) = 1 + 3, 2+2 = 2
P(4,3) = 1 + 1 + 2 = 1
P(4,4) = 1 + 1 + 1 + 1 = 1
C. How many ways can you distribute 4 identical balls among 4 different boxes
The number of ways in which 4 identical balls can be distributed among 4 different boxes = 
A. Yes ABCD is a trapezoid ONE way i remember this is trapezoids aerialist always the shape of some square/rectangle w/a triangle
B. You can always do tis on graph paper. So we will be moving point B from (1,5) to............... (-1,5)
Hope this helps!!!