Final answer:
In Mathematics, a panel of 12 jurors and 5 alternates can be chosen from a group of 40 prospective jurors in a specific number of ways calculated by using combinations C(40, 12) and C(28, 5), and then multiplying the two.
Step-by-step explanation:
The subject of this question is Mathematics, specifically combinatorics which is a topic often encountered in high school probability and statistics or algebra classes. The question asks us to determine the number of different ways a panel of 12 jurors and 5 alternates can be chosen from a group of 40 prospective jurors.
To solve this problem, we can use the concept of combinations since the order in which we select the jurors and alternates does not matter.
Firstly, we select 12 jurors from the 40 prospective jurors, which can be done in C(40, 12) ways, where C(n, k) is the combination of n items taken k at a time. After selecting the 12 jurors, we are left with 28 prospective jurors, from which we then select 5 alternates. This can be done in C(28, 5) ways. The two selections are independent of each other, so to find the total number of ways to select the jurors and alternates, we multiply the two combinations:
Total ways = C(40, 12) × C(28, 5)
To calculate the combinations, we can use the formula:
C(n, k) = n! / (k! × (n-k)!), where '!' denotes factorial.
Therefore, C(40, 12) would be calculated as:
40! / (12! × (40-12)!) = 40! / (12! × 28!)
And C(28, 5) would be:
28! / (5! × (28-5)!) = 28! / (5! × 23!)
=98280
Finally, to get the answer, we would calculate these two values and then multiply them.