# 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.

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### 2 Answers

The number ways we can buy 3 cows from six cows = 6C3.=6*5*4/3*2*1 =20

The number ways we can buy 2pigs from 5 pigs = 5C2 = 5*4/2*1 = 10

The number of ways we can buy 4 hens from 8hens = 8C4 = 8*7*6*5/4*3*2*1 = 70.

Since each of the activity is independent, we multiply the number of each of the ways that could be done = 20 *10*70 ways in all

= 14000 diffeent ways.

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**