Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 15 Jul 2019, 23:21

### GMAT Club Daily Prep

#### 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

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

# How many subordinates does Marcia have?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56234

### Show Tags

09 Jul 2013, 01:16
akijuneja wrote:
Bunuel wrote:
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.

Hi
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.

Posted from my mobile device

2^n is the number of lists including lists with 1 subordinate and an empty set (list with 0 subordinates is an empty set, which simply means that Marcia does not have any subordinate).

As for your other question, unfortunately I'm not sure I understand it.
_________________
Manager
Joined: 05 Nov 2012
Posts: 143

### Show Tags

26 Dec 2013, 21:40
$$200<2^n-n-1<500$$

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 $$289700<2^n-n-1<526100$$
Math Expert
Joined: 02 Sep 2009
Posts: 56234

### Show Tags

27 Dec 2013, 03:43
Amateur wrote:
$$200<2^n-n-1<500$$

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.
_________________
Intern
Joined: 10 Jan 2014
Posts: 22

### Show Tags

16 Feb 2014, 13:43
Bunuel wrote:
BarneyStinson wrote:
How many subordinates does Marcia Have?

(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.

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?
Math Expert
Joined: 02 Sep 2009
Posts: 56234

### Show Tags

17 Feb 2014, 07:43
damamikus wrote:
Bunuel wrote:
BarneyStinson wrote:
How many subordinates does Marcia Have?

(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.

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.
_________________
Intern
Joined: 20 Oct 2012
Posts: 7
Location: United States
Twilight: Sparkle
Concentration: Accounting, Economics
GPA: 3.85
Re: How many subordinates does Marcia have?  [#permalink]

### Show Tags

30 Apr 2014, 07:55
Hi Guys,
plz , I do not understand any of the answers , can anyone give a more simple explanation . Thank You in advance.
Math Expert
Joined: 02 Sep 2009
Posts: 56234
Re: How many subordinates does Marcia have?  [#permalink]

### Show Tags

30 Apr 2014, 07:57
gmatlover23 wrote:
Hi Guys,
plz , I do not understand any of the answers , can anyone give a more simple explanation . Thank You in advance.

It's a tough problem. Not every question has a silver bullet 20 sec solution.
_________________
Manager
Joined: 20 Oct 2013
Posts: 54

### Show Tags

12 May 2014, 05:52
Bunuel wrote:
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.

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
Manager
Joined: 28 Apr 2014
Posts: 205
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
Intern
Status: Preparing for GMAT
Joined: 10 Dec 2013
Posts: 14
Location: India
GMAT 1: 530 Q46 V18
WE: Other (Entertainment and Sports)
Re: How many subordinates does Marcia have?  [#permalink]

### Show Tags

26 Jul 2014, 13:14
From 1, how did we get n(n-1)/2! ?? Shouldn't it be n!/(n-2)!*2 ?? Can someone please explain!
Math Expert
Joined: 02 Sep 2009
Posts: 56234
Re: How many subordinates does Marcia have?  [#permalink]

### Show Tags

26 Jul 2014, 13:19
suhaschan wrote:
From 1, how did we get n(n-1)/2! ?? Shouldn't it be n!/(n-2)!*2 ?? Can someone please explain!

$$C^2_n$$;

$$\frac{n!}{(n-2)!*2!}$$;

$$\frac{(n-2)!*(n-1)*n}{(n-2)!*2!}$$;

$$\frac{n(n-1)}{2!}$$.

Does this make sense?
_________________
Re: How many subordinates does Marcia have?   [#permalink] 26 Jul 2014, 13:19

Go to page   Previous    1   2   [ 31 posts ]

Display posts from previous: Sort by

# How many subordinates does Marcia have?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne