BarneyStinson wrote:
How many subordinates does Marcia Have?
If
(1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates.
(2) there are 28 ways that she could decide which 2 subordinates she will recommend promoting.
This is a very very interesting problem and the best explanation I could come with is this -
Imagine you have nine open slots and nine digits from 1 to 9, repetition is allowed and you can fill those slots with the digits, it gives you 9^9 entire combinations.
Similarly, you have n open slots and n candidates, each candidate can either fill the slot or not. So that gives 2 possibilities for each slot. For n slots, 2^n. This is practically, all the possible combinations of filling in those slots.
Now given in stmt 1 is that - all such possible combinations range between 200 and 500. So 200 < 2^n < 500. Therefore, n = 8.
Little correction:
For (2) we have \(\{s_1,s_2,s_3,...s_n\}\). Each subordinate \((s_1,s_2,s_3,...s_n)\) has TWO options: either to be included in the list or not. Hence total # of lists - \(2^n\), correct. But this number will include \(n\) lists with 1 subordinate as well \(1\) empty list.
As the lists should contain
at least 2 subordinates, then you should subtract all the lists containing only 1 subordinate and all the lists containing 0 subordinate.
Lists with 1 subordinate - n: \(\{s_1,0,0,0...0\}\), \(\{0,s_2,0,0,...0\}\), \(\{0,0,s_3,0,...0\}\), ... \(\{0,0,0...s_n\}\).
List with 0 subordinate - 1: \(\{0,0,0,...0\}\)
So we'll get \(200<2^n-n-1<500\), --> \(n=8\). Sufficient.
For (1):
\(C^2_n=28\) --> \(\frac{n(n-1)}{2!}=28\) --> \(n(n-1)=56\) --> \(n=8\). Sufficient.
Answer: D.
I wanted to know how can the list be composed of zero subordinate and if there are other candidates as well who are not subordinate then answer will change.