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How many subordinates does Marcia have?
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16 Dec 2007, 02:13
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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.
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How many subordinates does Marcia have?
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25 Feb 2010, 17:24
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 as 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.
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Re: How many subordinates does Marcia have?
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30 Aug 2018, 07:56
marcodonzelli 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. aniket5552 reference your PM (1) There are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates. So atleast 2 means, she makes list of 2 people ,3 people ... till n people.. example say 5 people are there, so she makes a list of 2 out of 5 do 5C2, 3 out of 5 so 5C3, 4 out of 5 so 5C4 and 5 out of 5 so 5C5 people in a listHere let the number be n, so nC2+nC3+nC4+....nCn is what we are looking for... Now you should remember the formula nC0+nC1+....nCn=2^n So nC2+nC3+....nCn=2^nnC0nC1=2^nn1 This is between 200 and 500.... 2^7=128, so take next value 8....... 2^8=256 so 2^881=25681=247 Between 200 and 500 yes Take next value 2^991=51291=502.. outside the range Only 1 value that is 8 Sufficient (2) There are 28 ways that she could decide which 2 subordinates she will recommend promoting .. nC2=28......n(n1)=28*2......n(n1)=56=8*7 so n =8 Sufficient D
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Re: perms
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18 Dec 2007, 09:59
marcodonzelli 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.
I found it very hard..
For (1) remember that ii is often easier to count the converse of what is asked. Of the 2^n subsets of a list of n objects, how many contain less than 2 objects?




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B
1. Maybe I've not understood correctly but It seems to be insufficient.
2. nC2 describes this situation.
nC2=n(n1)/2=28 ==> n(n1)=56 ==> n=8



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really D.
first condition:
7: N=7C7+7C6+7C5+7C4+7C3+7C2=119<200
8: N=8C8+8C7+8C6+8C5+8C4+8C3+8C2= 238
9: N=9C9+9C8+9C7+9C6+9C5+9C4+9C3+9C2=504>500
but it is a terrible way. Any shortcuts?



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Re: perms
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16 Dec 2007, 12:08
marcodonzelli 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.
I found it very hard..
1: i am not sure how st 1 is sufficient as it says 200500 lists from at least 2 subordinates.
2: 28 ways = nc2 = n(n1)/2 = 8x7/2 = 28
so n = 8.
so is not it B?



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Re: perms
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18 Dec 2007, 11:45
kevincan wrote: 2^n
Thanks for that! It's really simple!



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Re: perms
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23 Dec 2007, 03:42
I actually didn't catch this expalantion..I have had another explanation as well, but even this one is not clear to me.... let's exclude from the 2^n lists, the only that have only one name and the 1 that has none, so 200<2^n  n  1<500. so we have n=8.
Why do we have to consider 2^n lists? what's that mean: the n that have only one name and the one that has none...please help me!



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Re: perms
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23 Dec 2007, 18:31
marcodonzelli wrote: I actually didn't catch this expalantion..I have had another explanation as well, but even this one is not clear to me.... let's exclude from the 2^n lists, the only that have only one name and the 1 that has none, so 200<2^n  n  1<500. so we have n=8.
Why do we have to consider 2^n lists? what's that mean: the n that have only one name and the one that has none...please help me!
nc0 + nc1 + nc2 + nc3 + ..... + ncn = 2^n
How? If you are familiar with Binomial expansion of (a + b)^n, substitute a=1 and b=1
http://www.mathwords.com/b/binomial_theorem.htm



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Re: perms
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24 Dec 2007, 06:34
marcodonzelli 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.
I found it very hard..
2) is absolutely clear.
1)...I am not even able to comprehend the statement. I am not familiar with binomials, could someone please explain the "simple logic" behind 2^n.
I do understand that...... 2^n gives the number of combinations possible for n things.
But what does it have to do with lists? what is list implying in the question? and number of lists are not fixed in the question... it says it is varying from 200 to 500 ( in case we assume that 200 and 500 are the number of combinations)
Please help!



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Re: perms
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24 Dec 2007, 07:20
LM wrote: 1)...I am not even able to comprehend the statement. I am not familiar with binomials, could someone please explain the "simple logic" behind 2^n.
I do understand that...... 2^n gives the number of combinations possible for n things.
But what does it have to do with lists? what is list implying in the question? and number of lists are not fixed in the question... it says it is varying from 200 to 500 ( in case we assume that 200 and 500 are the number of combinations)
Please help!
maybe my logic will help you.
1. we have n subordinates: S={1,2,3,4....,n}
2. each subordinate may be included or not be included in a list. ( two possibilities)
3. our list we can image like a={1,0,0,1,1,0,1,0,....1}  where 1  in the list, 1  out of the list.
4. How many lists we can compose? N=2*2*2......2=2^n
5. N is not any number and is from set: {2^1,2^2,2^3,...,2^n,...}
6. Now, we have 2^7=128,2^8=256,2^9=512
7. we should exclude n(1subordinate lists)+1(empty list)
8. 120, 247,502



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Re: perms
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14 Apr 2008, 17:14
Boy. This was cool. I guess just reading answer gave me some cool missing insights. Sounds tough for exam though. Hope, now I get it Thanks all contributors.



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Re: perms
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19 Oct 2009, 23:45
The answer has to be B and not D as stated. Since a range is mentioned in A. when there is a range multiple answers arise which is against DS. Hence B, unless there is some other explaination. Please provide source of the question, to judge the questions authencity



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Re: perms
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29 Nov 2009, 00:54
sapbi wrote: The answer has to be B and not D as stated. Since a range is mentioned in A. when there is a range multiple answers arise which is against DS. Hence B, unless there is some other explaination. Please provide source of the question, to judge the questions authencity You are right for most cases, but here only n = 8 fits in the range of 200 to 500, so answer is D.



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Re: perms
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25 Feb 2010, 14:45
the easiest way i cld think of was to draw a pascals triangle first 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 ..... 256 1 9 36 84 126 126 84 36 9 1 ..... 512
(i) adding all the elements in the individual row, only the last before row fall in the 200 to 500 category, so n=8
(ii) match nC2 = 28 in the above triangle and the only row that satisfies is last before row. so n=8
So, D is the answer
Shashi



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Re: perms
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25 Feb 2010, 16:49
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.



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25 Feb 2010, 18:19
[quote="Bunuel"][/quote] I feel so humbled that you care for me, Math God!!!



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Re: How many subordinates does Marcia Have? 1) there are between
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28 Nov 2011, 08:35
wOOW... I just learnt about Binomials from this thread thanks guys



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Re: perms
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09 Jul 2013, 00:02
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 \(200<2^nn1<500\), > \(n=8\). Sufficient. For (1): \(C^2_n=28\) > \(\frac{n(n1)}{2!}=28\) > \(n(n1)=56\) > \(n=8\). Sufficient. Answer: D. 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. Please clarify Posted from my mobile device







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