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How many subordinates does Marcia have? 16 Dec 2007, 03:13
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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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Bunuel
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How many subordinates does Marcia have? 25 Feb 2010, 18:24
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BarneyStinson
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.
Show more

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

Answer: D.
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How many subordinates does Marcia have? 24 Dec 2007, 08:20
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LM
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!
Show more


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(1-subordinate lists)+1(empty list)
8. 120,247,502
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How many subordinates does Marcia have? 18 Dec 2007, 10:59
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marcodonzelli
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..
Show more


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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How many subordinates does Marcia have? 23 Dec 2007, 04:42
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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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How many subordinates does Marcia have? 23 Dec 2007, 19:31
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marcodonzelli
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!
Show more


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
https://www.mathwords.com/b/binomial_theorem.htm
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How many subordinates does Marcia have? 24 Dec 2007, 07:34
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marcodonzelli
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..
Show more


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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How many subordinates does Marcia have? 25 Feb 2010, 17:49
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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.
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How many subordinates does Marcia have? 25 Feb 2010, 19:19
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[quote="Bunuel"][/quote]

I feel so humbled that you care for me, Math God!!! :-) :-)
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How many subordinates does Marcia have? 09 Jul 2013, 01:02
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BarneyStinson
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.

Answer: D.
Show more

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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How many subordinates does Marcia have? 09 Jul 2013, 01:16
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BarneyStinson
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.

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
Show more

2^n is the number of lists including lists with 1 subordinate and an empty set (list with 0 subordinates is an empty set, which simply means that Marcia does not have any subordinate).

As for your other question, unfortunately I'm not sure I understand it.
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How many subordinates does Marcia have? 26 Jul 2014, 13:14
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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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How many subordinates does Marcia have? 26 Jul 2014, 13:19
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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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\(C^2_n\);

\(\frac{n!}{(n-2)!*2!}\);

\(\frac{(n-2)!*(n-1)*n}{(n-2)!*2!}\);

\(\frac{n(n-1)}{2!}\).

Does this make sense?
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How many subordinates does Marcia have? 30 Aug 2018, 08:56
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marcodonzelli
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.
Show more


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 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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How many subordinates does Marcia have? 21 Oct 2022, 07:57
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marcodonzelli
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.
Show more


Wow, I understood this question in a completely different manner :
I understood it was about the number of different lists Marcia could form using the names of her subordinates,
so something like "arranging x number of names".

Statement 1) would give : 200 < # arrangements of x names < 500

-> arranging 5 names : 5! = 120
-> arranging 6 names : 6! = 720
-> arranging 6 names with 2 subordinates sharing the same name : 6!/2! = 360 => Bingo
Therefore there are in total 6 subordinates, sufficient.

Statement 2) would just be solving :
x!/x!(x-2)! = 28
x(x-1) = 56
x = 8
sufficient
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How many subordinates does Marcia have? 20 May 2025, 06:45
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