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# How many subordinates does Marcia have?

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Joined: 02 Sep 2009
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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.
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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 $$289700<2^n-n-1<526100$$
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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.
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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?
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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.
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Re: How many subordinates does Marcia have?  [#permalink]

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

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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.
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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?
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How many subordinates does Marcia have?  [#permalink]

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

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

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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?
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Re: How many subordinates does Marcia have?   [#permalink] 26 Jul 2014, 13:19

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