Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

I did get this equation but how do you solve it guys? GMAT may not be testing it but I am interested to know the process of solving this equation. This one is easy by plugging, but what if the equation is \(289700<2^n-n-1<526100\)

I did get this equation but how do you solve it guys? GMAT may not be testing it but I am interested to know the process of solving this equation. This one is easy by plugging, but what if the equation is \(289700<2^n-n-1<526100\)

You won't be asked to solve \(289700<2^n-n-1<526100\) on the GMAT.
_________________

(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'm sorry for bringing this old post up again but I'm afraid I don't quite understand your solution.

For statement 1, why did you use 2^n here and not a factorial approach? I just don't understand the logic behind it. For example, if Marcia had 4 different subordinates - e.g Anna, Bea, Christian, Doug (ABCD) - then there would be 4!/(2!2!) ways to create a list with two employee names on it. if the number of different two-name lists is to be between 200 and 500, then Marcia could have either 21 employees (21!/(2!19!)=210) or 32 employees (32!/2!30!=496). Since we do not know how many different employee names she puts on those lists, we can't really tell how many different subordinates she has.

I know I made a mistake somewhere, but I don't know where or why my approach/ logic is wrong. Could you please enlighten me?

(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'm sorry for bringing this old post up again but I'm afraid I don't quite understand your solution.

For statement 1, why did you use 2^n here and not a factorial approach? I just don't understand the logic behind it. For example, if Marcia had 4 different subordinates - e.g Anna, Bea, Christian, Doug (ABCD) - then there would be 4!/(2!2!) ways to create a list with two employee names on it. if the number of different two-name lists is to be between 200 and 500, then Marcia could have either 21 employees (21!/(2!19!)=210) or 32 employees (32!/2!30!=496). Since we do not know how many different employee names she puts on those lists, we can't really tell how many different subordinates she has.

I know I made a mistake somewhere, but I don't know where or why my approach/ logic is wrong. Could you please enlighten me?

The lists should contain at least 2 subordinates, not exactly 2.
_________________

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.

Dear Bunnel

How can you even understand what the qs is saying... i couldnt understand what statement 1 was saying? what list are they talking about?
_________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

How many subordinates does Marcia have? [#permalink]

Show Tags

06 Jul 2014, 23:49

BarneyStinson wrote:

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.

Just out of interest , how will the above understanding change if we have say 'a' open slots and 'b' candidates ( b>a) ?. Will it be bCa (2^a) ?

Bunuel any pointers for more information on this concept will be much appreciated

Re: How many subordinates does Marcia have? [#permalink]

Show Tags

19 Oct 2017, 22:58

How did you guys thought of using 2^n rather than nC2+nC3+nC4...+nCn. I started by assuming she has 4 subordinates, then i concluded A is not sufficient since we dont know n. Kindly enlighten me on where to use 2^n and where to use nCr approach.?