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Re: perms [#permalink]
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26 Dec 2013, 21:40
\(200<2^nn1<500\) 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 \(289700<2^nn1<526100\)



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Re: perms [#permalink]
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27 Dec 2013, 03:43



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Re: perms [#permalink]
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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 \(200<2^nn1<500\), > \(n=8\). Sufficient. For (1): \(C^2_n=28\) > \(\frac{n(n1)}{2!}=28\) > \(n(n1)=56\) > \(n=8\). 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 twoname 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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Re: perms [#permalink]
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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 \(200<2^nn1<500\), > \(n=8\). Sufficient. For (1): \(C^2_n=28\) > \(\frac{n(n1)}{2!}=28\) > \(n(n1)=56\) > \(n=8\). 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 twoname 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



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Re: perms [#permalink]
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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 \(200<2^nn1<500\), > \(n=8\). Sufficient.
For (1): \(C^2_n=28\) > \(\frac{n(n1)}{2!}=28\) > \(n(n1)=56\) > \(n=8\). 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?
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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(n1)/2! ?? Shouldn't it be n!/(n2)!*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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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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