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# How many subordinates does Marcia have?

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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
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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
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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]

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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.?

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Re: How many subordinates does Marcia have?   [#permalink] 19 Oct 2017, 22:58

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# How many subordinates does Marcia have?

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