Final answer:
To assign 18 accounts to three salespeople, we calculate the combinations for 6 accounts out of 18 for the first person and then 6 out of the remaining 12 for the second. No further combinations are needed for the third. The total ways to assign the accounts is the product of these two combination values.
Step-by-step explanation:
The subject of this question is combinatorics, a branch of mathematics concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. The question asks how to distribute 18 accounts among three salespeople: Joe, Megan, and Santana, with each receiving six accounts.
To solve this, we can use the concept of combinations. Since the order of selection doesn't matter, we want to find how many ways we can choose 6 accounts out of 18 for the first person, then 6 out of the remaining 12 for the second person, and the rest will go to the third person.
The number of ways to choose 6 accounts out of 18 for the first person is given by the combination formula C(n, k) = n! / (k!(n - k)!), where n is the total number of items, and k is the number of items to choose. So, for the first person, it's C(18, 6). After assigning six accounts to the first person, we have 12 accounts left.
Then, we choose 6 out of these 12 for the second person, which is C(12, 6). The remaining 6 accounts will automatically go to the third person, so no further combinations are needed here.
The total number of ways to distribute the accounts is thus C(18, 6) × C(12, 6). To find the numerical values, we would calculate:
- C(18, 6) = 18! / (6! × (18 - 6)!) = 18,564
- C(12, 6) = 12! / (6! × (12 - 6)!) = 924
Thus, the total ways to assign the accounts is 18,564 × 924 = 17,154,576 ways.