In how many different ways can a panel of 12 jurors and 2 alternates be chosen from a group of 15 prospective jurors?

If there are 14 people, there are

14C2 ways to choose the 2 alternatives.

If there are 15 people, there are
15C14 ways to choose the 14 people.

So the number of ways of choosing 12 jurors and 2 alternatives is
15C14*14C2

To find the number of different ways a panel of 12 jurors and 2 alternates can be chosen from a group of 15 prospective jurors, we can use the concept of combinations.

In this problem, we need to choose 12 jurors out of 15 prospective jurors. The order in which we choose the jurors does not matter, so we can use the concept of combinations, specifically the combination formula.

The combination formula is given by:

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

where n is the total number of items, k is the number of items to be chosen, and the exclamation mark (!) denotes the factorial of a number, which is the product of all positive integers less than or equal to that number.

Applying the combination formula to this problem, we have:

C(15, 12) = 15! / (12! * (15 - 12)!)

Simplifying the factorial terms, we get:

C(15, 12) = 15! / (12! * 3!)

Now, calculate the factorials:

15! = 15 * 14 * 13 * 12!

3! = 3 * 2 * 1

Substituting the factorials back into the formula, we have:

C(15, 12) = (15 * 14 * 13 * 12!) / (12! * 3 * 2 * 1)

Simplifying the common terms in the numerator and denominator, we get:

C(15, 12) = (15 * 14 * 13) / (3 * 2 * 1)

Evaluating the right side of the equation, we find:

C(15, 12) = 455

Therefore, there are 455 different ways a panel of 12 jurors and 2 alternates can be chosen from a group of 15 prospective jurors.