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Director  Joined: 22 Nov 2007
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How many subordinates does Marcia have?  [#permalink]

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Question Stats: 22% (02:10) correct 78% (01:44) wrong based on 1197 sessions

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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?  [#permalink]

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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 $$200<2^n-n-1<500$$, --> $$n=8$$. Sufficient.

For (1):
$$C^2_n=28$$ --> $$\frac{n(n-1)}{2!}=28$$ --> $$n(n-1)=56$$ --> $$n=8$$. Sufficient.

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

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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.

(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 list
Here 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^n-nC0-nC1=2^n-n-1
This is between 200 and 500....
2^7=128, so take next value 8.......
2^8=256 so 2^8-8-1=256-8-1=247 Between 200 and 500 yes
Take next value 2^9-9-1=512-9-1=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(n-1)=28*2......n(n-1)=56=8*7 so n =8
Sufficient

D
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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?
##### General Discussion
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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(n-1)/2=28 ==> n(n-1)=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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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 200-500 lists from at least 2 subordinates.

2: 28 ways = nc2 = n(n-1)/2 = 8x7/2 = 28
so n = 8.

so is not it B?
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kevincan wrote:
2^n

Thanks for that! It's really simple! Director  Joined: 22 Nov 2007
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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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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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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)

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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)

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(1-subordinate lists)+1(empty list)
8. 120,247,502
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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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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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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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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

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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.
Senior Manager  Joined: 21 Jul 2009
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[quote="Bunuel"][/quote]

I feel so humbled that you care for me, Math God!!!  Manager  Status: Essaying
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Re: How many subordinates does Marcia Have? 1) there are between  [#permalink]

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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 Re: perms   [#permalink] 09 Jul 2013, 00:02

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