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in how many ways can a member of the committee select 5 applicants, ranking their first and second choices, and providing 3 unranked alternates?

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Final answer:

There are 203,222,800 ways in which a member of the committee can select 5 applicants, rank their first and second choices, and provide 3 unranked alternates.

Step-by-step explanation:

To determine the number of ways a member of the committee can select 5 applicants and rank their first and second choices, and provide 3 unranked alternates, we can use the concept of permutations.

The number of ways to select 5 applicants from the total pool is denoted by 10P5, which can be calculated as:

10P5 = (10!)/(10-5)! = (10!)/(5!) = (10 × 9 × 8 × 7 × 6)/(5 × 4 × 3 × 2 × 1) = 30240

Next, we need to consider the ranking of the first and second choices. Since each of the selected 5 applicants can be assigned either as the first choice or the second choice, we need to multiply the previous result by 2^5:

30240 × 2^5 = 30240 × 32 = 967,680

Finally, we need to consider the 3 unranked alternates. Each of the 3 unranked alternates can be selected from the remaining pool of applicants, so we multiply the previous result by the number of ways to select 3 from the remaining applicants:

967,680 × 7P3 = 967,680 × (7!)/(7-3)! = 967,680 × (7!)/(4!) = 967,680 × (7 × 6 × 5) = 967,680 × 210 = 203,222,800

Therefore, there are 203,222,800 ways in which a member of the committee can select 5 applicants, rank their first and second choices, and provide 3 unranked alternates.

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