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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]
06 Jul 2014, 22:49

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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