These are three questions with three complete answers.
Answers:
(a) C(100,6) = 100! / [ 9! × (100 -9)! ] =
= (100×99×98×97×96×95×94×93×92) / (9×8×7×6×5×4×3×2×1) =
= 1,902,231,808,400
(b) C(9,6) = 9! / [ 6! * (9 - 6)! ] = 9! / [6! 3!] = (9 × 8 × 7 × 6!) (6! × 3 × 2 × 1) =
= (9 × 8 × 7 × 6!) (6! × 3 × 2 × 1) = (9 × 8 × 7 ) / (3 × 2 × 1) = 84
(c) P(6,3) = 6! / (6 - 3)! = 6! / 3! = (6 × 5 × 4 × 3!) / 3! = 120
Step-by-step explanation:
(a) If 100 applicants apply for the job, how many ways are there to select a subset of 9 for a short list?
This is the formula for combinations: C (m,n) = m! / [n! (m - n)! ].
We will also use the formula for permutations, only as an intermediate step, to explain the solution. The formula for permutations is: P (m,n) = m! / (m - n)!
Next you will see why the final formula that you can use to solve the problem is that of combinations (because the order in which you make the list does not matter) and how you use it.
You have to select a subset of 9 candidates from a list of 100 applicants.
The first candidate may be chosen from the 100 different applicants, the second candidate may be chosen from the 99 left applicants, the third candidate from 98 applicants, and so on, which leads to:
- 100 × 99 × 98 × 97 × 96 × 95 × 94 × 93 × 92 possible variants.
Note that this is the permutation of 100 candidates taken from 9 in 9:
P(100,9) = 100! (100 - 9)! = 100! / (91!) =
= 100 × 99 × 98 × 97 × 96 × 95 × 94 × 93 × 92 × 91! / 91! =
= 100× 99 × 98 × 97 × 96 × 95 × 94 × 93 × 92.
But you have to eliminate the repetitions!
Suppose that A, B, C, D, E, F, G, H, I represents the set formed by nine selected members whose names are A, B, C, D, E, F, G, H and I. So, any combination of those same names, written in different order, represents the same set (list). That means that there are 9! = 9× 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 equivalent lists.
That is why you must divide the first result (possible ways in which you can select nine candidates) by the number of ways that represent the same list for every set.
So, the conclusion is that the number of different lists of nine candidates is:
C(100,6) = 100! / [ 9! × (100 -9)! ] =
= (100×99×98×97×96×95×94×93×92) / (9×8×7×6×5×4×3×2×1) =
= 1,902,231,808,400
(b) If 6 of the 9 are selected for an interview, how many ways are there to pick the set of people who are interviewed? (You can assume that the short list is already decided).
Since, the short list, i.e. the subset of 9 candidates is already decided, you will select 6 candidates to interview from 9 possible candidates.
So, your final set of candidates to interview will be the combination of 9 candidates taken from 6 in 6. The order of the names A, B, C, D, E, F, and G, is not relevant, and, therefore, the formula to use is that of combinations:
- C (m,n) = m! / [n! (m - n)! ]
- C(9,6) = 9! / [ 6! * (9 - 6)! ] = 9! / [6! 3!] = (9 × 8 × 7 × 6!) (6! × 3 × 2 × 1) =
= (9 × 8 × 7 × 6!) (6! × 3 × 2 × 1) = (9 × 8 × 7 ) / (3 × 2 × 1) = 84
(c) Based on the interview, the committee will rank the top three candidates and submit the list to their boss who will make the final decision. (You can assume that the interviewees are already decided.) How many ways are there to select the list from the 6 interviewees?
Ranking the top three candidates means that the order matters. Because it is not the same A, B, C than A, C, B, nor B, A, C, nor B, C, A, nor C, A, B, nor C, A, B.
Hence, you have to use the formula for permutations (not combinations).
The formula is: P(m,n) = m! / (m - n)!
Here, you must rank (select) 3 names, from a set (list) of 6 names, and the formula yields to:
- P(6,3) = 6! / (6 - 3)! = 6! / 3! = (6 × 5 × 4 × 3!) / 3! = 120
