We want to find different ways of selecting and ordering elements. The answers are:
A) C = 395
B) combinations = 479,001,600
C) combinations = 239,500,800
First, we know that Rebecca has 12 photos.
A) She wants to choose 8 of these 12 photos, we want to see in how many ways she can choose these 8 photos.
Remember that for a set of N elements, the number of different combinations of K elements (such that K is smaller than N) is given by:
[tex]C(N, K) = \frac{N!}{(N - K)!*K!}[/tex]
Here we have N = 12 and K = 8, then:
[tex]C(12, 8) = \frac{12!}{(12 - 8)!*8!} = \frac{12!}{4!*8!} = \frac{12*11*10*9}{4*3*2*1} = 495[/tex]
So there are 495 different ways of choosing 8 photos out of the 12.
B) Now we want to see in how many ways she can arrange all 12 photos.
The total number of combinations is just the product between the numbers of options, we got:
[tex]combinations = 12*11*10*9*8*7*6*5*4*3*2*1 = 479,001,600[/tex]
C) Now we must combine the two things we did before, first, we select 10 out of the 12 photos, and then we want to find in how many ways we can arrange these 10 photos.
[tex]C(12, 10) = \frac{12!}{(12 - 10)°*10°} = \frac{12*11}{2} = 66[/tex]
And we multiply this by the total number of arrangements of these 10 photos, that is just given by 10!, so we get:
[tex]c = 66*10! = 66*(10*9*8*7*6*5*4*3*2*1) = 239,500,800[/tex]
If you want to learn more, you can read:
https://brainly.com/question/14305145
