# 10) A jury pool of consists of 50 potential jurors. In how many ways can a jury of 12 be selected?

## 10) A jury pool of consists of 50 potential jurors. In how many ways can a jury of 12 be selected?

Apr 16, 2018

""_50C_12 ways.

(""_50C_12 = 121,399,651,100)

#### Explanation:

We have:

50 choices for the first member,
49 choices for the second member,
....etc.
40 choices for the 11th member, and
39 choices for the 12th member.

If we want to keep the jurors in the order we picked them, we have

$\textcolor{w h i t e}{=} 50 \times 49 \times 48 \times \ldots \times 40 \times 39$

=(50!)/(38!)

=(50!)/((50-12)!)

ways to create an ordered jury.

This can be written as ""_50P_12 to indicate we are permuting 12 units from a population of 50 (order matters).

But, since order does not matter in this case (i.e. all jurors have the same "rank"), we need to divide this by the number of ways these 12 jurors can be ordered. That number is 12!.

The number of juries possible is then:

color(white)= (50!)/(38!)-:12!

=(50!)/((50-12)!xx12!)

This can be written as ""_50C_12 to indicate we are combining 12 units from a population of 50 (order does not matter).

Apr 16, 2018

$121 , 399 , 651 , 100$

#### Explanation:

Classic combinatorics problem. To enumerate the number of ways $k$ items can be chosen from $n$ total items we calculate

(n!)/(k!(n-k)!)=((50!)/(12!*38!))

$= \frac{50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43 \cdot 42 \cdot 41 \cdot 40 \cdot 39}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}$

Prime factorization and cancelling show that this is equal to

$47 \cdot 43 \cdot 41 \cdot 23 \cdot 13 \cdot {7}^{2} \cdot {5}^{2} \cdot {2}^{2} = 121 , 399 , 651 , 100$