A farmer buys three cows , two pigs and for hens from a man who has six cows , five pigs and eight hens. Determine the choices that the farmer have.

To determine the number of choices of the farmer, we'll
apply combinations.

We'll recall the formula of the
combination of n elements taken k at a
time:

C(n,r) =
n!/k!(n-k)!

So the farmer has the
choices:

C(6,3)*C(5,2)*C(8,4)

We'll calculate
the combinations:

C(6,3) =
6!/3!(6-3)!

C(6,3) =
3!*4*5*6/3!*1*2*3

We'll simplify and we'll
get:

C(6,3) =
4*5

C(6,3) =
20

C(5,2) =
5!/2!(5-2)!

C(5,2)
= 5!/2!3!

C(5,2) =
3!*4*5/1*2*3!

C(5,2) =
2*5

C(5,2) =
10

C(8,4) =
8!/4!(8-4)!

C(8,4) =
8!/4!4!

C(8,4) =
4!5*6*7*8/4!1*2*3*4

C(8,4) =
5*7*2

C(8,4) =
70

The number ways the farmer can choose
is:

C(6,3)*C(5,2)*C(8,4) =
20*10*70

C(6,3)*C(5,2)*C(8,4) = 14000
choices