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Company X is forming a two-person committee by selecting one

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Company X is forming a two-person committee by selecting one  [#permalink]

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New post Updated on: 11 Jul 2013, 00:35
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Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.

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Originally posted by martie11 on 20 Sep 2011, 20:12.
Last edited by Bunuel on 11 Jul 2013, 00:35, edited 5 times in total.
Edited the question.
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Re: Company X is forming a two-person committee by selecting one  [#permalink]

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New post 21 Mar 2014, 01:40
1
5
vishalrastogi wrote:
martie11 wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.


Bunuel - Pl help, not able to understand the solution !

Thanks.


Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}.

The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.

Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member:
\(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\)

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient.

(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.

(1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient:
x=1, y=30, z=5;
x=2, y=15, z=5;
x=3, y=10, z=5;
x=5, y=6, z=5;
x=6, y=5, z=5;
x=10, y=3, z=5;
x=15, y=2, z=5;
x=30, y=1, z=5.

Answer: E.

Hope it's clear.
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Re: 700+ DS...Company X is forming a two-person committee...  [#permalink]

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New post 20 Sep 2011, 23:57
1
martie11 wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

i) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
ii) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.


Using both statements:

Case I:
M Only:2, B only:15, M&B:10

2*15=30 Satisfies
Both M&B=10 Satisfies

So,
\(Committees=C^{10}_1*C^{15}_1+C^{2}_1*C^{10}_1+C^{10}_1*C^{9}_1=150+10+90=250\)

Case II:
M Only:30, B only:1, M&B:10

1*30=30 Satisfies
Both M&B=10 Satisfies

So,
\(Committees=C^{10}_1*C^{30}_1+C^{1}_1*C^{10}_1+C^{10}_1*C^{9}_1=300+10+90=400\)

Not Sufficient.

We could have taken other examples as well:
{6,5} as 6*5=30
{3,10} as 3*10=30

Ans: "E"
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Re: 700+ DS...Company X is forming a two-person committee...  [#permalink]

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New post 21 Sep 2011, 00:32
1
to answer this question, the big question is how many managers are there in the company X and the form: P(m)+P(b)-P(m&b)-None
statement 1 lets you know none is 30 surely insufficient
statement 2 lets you know both P(m&b)= 10 insufficient
both cant give you the answer of the question how many manahers there are in the company X
so E
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Re: 700+ DS...Company X is forming a two-person committee...  [#permalink]

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New post 21 Sep 2011, 22:33
managers =m , board members = b and common member = mb

now required combination = (m+b)C2 - (m-mb)C1*(b-mb)C1


a (m-mb)*(b-mb)= 30 hence not sufficient.

b mbC2 = 10 meaning - mb*(mb-1)/2 = 10 thus mb = 5

not sufficient.

a+b (m-5)*(b-5) = 30 meaning m,b can be 20,7 or 15,8. hence not sufficient.

thus E it is.
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Re: 700+ DS...Company X is forming a two-person committee...  [#permalink]

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New post 29 Sep 2011, 08:43
martie11 wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

i) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
ii) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

Thank you.


E should be the OA since we do not know the number of managers and board members
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Re: Company X is forming a two-person committee by selecting one  [#permalink]

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New post 17 Oct 2013, 02:31
martie11 wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.


I know everyone got it right but there is more to this question .. this is a good question ..

here:
no. of managers - M
Board members - B

it can be represented in a venn Diagram ...
a+c = M(a - only managers, c - managers as well as board member)
b+c = B)b - only board members, c - manafers as well as board members)

statement1:
aC1.bC2 = 30(selected one manager and one board member)

a.b = 60 .. now there are many combinations for a.b such as 1*60, 2*30, 3*20 ...... hence a+b can be 61,32,23 ..... not sufficient

statement2:
cC2=10 (selected 2 persons which are both manager and board member)
c(c-1) = 20 .. only possible answer for c is 5 ..



... If the question would've asked no. of members which are both manager and board member we could've answered it with B.
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Re: Company X is forming a two-person committee by selecting one  [#permalink]

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New post 08 May 2014, 04:52
1
Bunuel wrote:
vishalrastogi wrote:
martie11 wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.


Bunuel - Pl help, not able to understand the solution !

Thanks.


Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}.

The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.

Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member:
\(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\)

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient.

(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.

(1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient:
x=1, y=30, z=5;
x=2, y=15, z=5;
x=3, y=10, z=5;
x=5, y=6, z=5;
x=6, y=5, z=5;
x=10, y=3, z=5;
x=15, y=2, z=5;
x=30, y=1, z=5.

Answer: E.

Hope it's clear.


Bunuel a couple of doubts over here.

When you say

The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.


Why is z different ? Won't it be made from x,y ?

Also not clear on why the probability aspects came into your solution ? Isn't this a question of combinations ?
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Re: Company X is forming a two-person committee by selecting one  [#permalink]

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New post 08 May 2014, 07:12
himanshujovi wrote:
Bunuel wrote:
vishalrastogi wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}.

The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.

Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member:
\(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\)

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient.

(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.

(1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient:
x=1, y=30, z=5;
x=2, y=15, z=5;
x=3, y=10, z=5;
x=5, y=6, z=5;
x=6, y=5, z=5;
x=10, y=3, z=5;
x=15, y=2, z=5;
x=30, y=1, z=5.

Answer: E.

Hope it's clear.


Bunuel a couple of doubts over here.

When you say

The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.


Why is z different ? Won't it be made from x,y ?

Also not clear on why the probability aspects came into your solution ? Isn't this a question of combinations ?


Managers and board members form overlapping sets.
Attachment:
Untitled.png
Untitled.png [ 4.91 KiB | Viewed 4604 times ]

The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.
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Collection of Questions:
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Re: Company X is forming a two-person committee by selecting one  [#permalink]

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New post 10 May 2014, 00:53
Bunuel wrote:
vishalrastogi wrote:
martie11 wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.


Bunuel - Pl help, not able to understand the solution !

Thanks.


Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}.

The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.

Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member:
\(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\)

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient.

(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.

(1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient:
x=1, y=30, z=5;
x=2, y=15, z=5;
x=3, y=10, z=5;
x=5, y=6, z=5;
x=6, y=5, z=5;
x=10, y=3, z=5;
x=15, y=2, z=5;
x=30, y=1, z=5.

Answer: E.

Hope it's clear.



Hello Bunuel

Question Asks us to find the Number of Committees, You have given you solutn by finding the probability.
Although doest makes much difference !
I tried a slight different method. Plz tell me where i am going wrong ?

Sorry for wasting your time :(

The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.

Let t= x+y.
t shows the total number of members other than the one who are both Managers & Board Members.

We have to find out: How many committees can be formed such that at least one committee member is both a manager and a board member.

So it can be := \(C^1_z . C^1_t + C^2_z\)

where \(C^1_z . C^1_t\) = one member from the members who are both Director & board member, Other from the remaining i.e. t

\(C^2_z\) = both memebers from the members who are both Director & board member.


(1)
The number of committees that can be formed such that neither committee member is both a manager and a board member is 30
t represents the total of neither

so we get : \(C^2_t = 30\) --> \(\frac{(t-1)t}{2}=30\)

Not sufficient.


(2)
The number of committees that can be formed such that each committee member is both a manager and a board member is 10

--> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.

(Same as yours)


1+2)
We had to find : \(C^1_z . C^1_t + C^2_z\)

From eq.2 We know Z
from eq.1 It seems we can find out t - although its not giving integer values. I dont knw why.

This way i was getting C as the answer

Please help
Thanks a lot Again !! :)
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Re: Company X is forming a two-person committee by selecting one  [#permalink]

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New post 10 May 2014, 05:21
1
niyantg wrote:
Bunuel wrote:
vishalrastogi wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}.

The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.

Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member:
\(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\)

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient.

(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.

(1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient:
x=1, y=30, z=5;
x=2, y=15, z=5;
x=3, y=10, z=5;
x=5, y=6, z=5;
x=6, y=5, z=5;
x=10, y=3, z=5;
x=15, y=2, z=5;
x=30, y=1, z=5.

Answer: E.

Hope it's clear.



Hello Bunuel

Question Asks us to find the Number of Committees, You have given you solutn by finding the probability.
Although doest makes much difference !
I tried a slight different method. Plz tell me where i am going wrong ?

Sorry for wasting your time :(

The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.

Let t= x+y.
t shows the total number of members other than the one who are both Managers & Board Members.

We have to find out: How many committees can be formed such that at least one committee member is both a manager and a board member.

So it can be := \(C^1_z . C^1_t + C^2_z\)

where \(C^1_z . C^1_t\) = one member from the members who are both Director & board member, Other from the remaining i.e. t

\(C^2_z\) = both memebers from the members who are both Director & board member.


(1)
The number of committees that can be formed such that neither committee member is both a manager and a board member is 30
t represents the total of neither

so we get : \(C^2_t = 30\) --> \(\frac{(t-1)t}{2}=30\)

Not sufficient.


(2)
The number of committees that can be formed such that each committee member is both a manager and a board member is 10

--> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.

(Same as yours)


1+2)
We had to find : \(C^1_z . C^1_t + C^2_z\)

From eq.2 We know Z
from eq.1 It seems we can find out t - although its not giving integer values. I dont knw why.

This way i was getting C as the answer

Please help
Thanks a lot Again !! :)


\(C^2_t = 30\) in your solution includes the cases when both members are managers as well as the cases when both members are board members, while we are told that one member must be from the managers and the other member from the board members. So, the first statement should be translated as \(x*y=30\).

Does this make sense?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Company X is forming a two-person committee by selecting one  [#permalink]

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New post 11 May 2014, 21:56
Quote:
\(C^2_t = 30\) in your solution includes the cases when both members are managers as well as the cases when both members are board members, while we are told that one member must be from the managers and the other member from the board members. So, the first statement should be translated as \(x*y=30\).

Does this make sense?


Ohhhh !
I fotgot to take into consideration the condition given in the Question.
Thanksss A lot Bunuel

You are GREAT ! :) :)
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Re: Company X is forming a two-person committee by selecting one  [#permalink]

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New post 12 May 2014, 03:00
Bunuel wrote:
vishalrastogi wrote:
martie11 wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. Howmany committees can be formed such that at least one committee member is both a manager and a board member?

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.


Bunuel - Pl help, not able to understand the solution !

Thanks.


Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}.

The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.

Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member:
\(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\)

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient.

(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.

(1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient:
x=1, y=30, z=5;
x=2, y=15, z=5;
x=3, y=10, z=5;
x=5, y=6, z=5;
x=6, y=5, z=5;
x=10, y=3, z=5;
x=15, y=2, z=5;
x=30, y=1, z=5.

Answer: E.

Hope it's clear.


Sorry for being silly but are we asked probability or total number of combinations ?
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Re: Company X is forming a two-person committee by selecting one  [#permalink]

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New post 12 May 2014, 03:13
himanshujovi wrote:
Bunuel wrote:
vishalrastogi wrote:

Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}.

The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.

Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member:
\(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\)

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient.

(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.

(1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient:
x=1, y=30, z=5;
x=2, y=15, z=5;
x=3, y=10, z=5;
x=5, y=6, z=5;
x=6, y=5, z=5;
x=10, y=3, z=5;
x=15, y=2, z=5;
x=30, y=1, z=5.

Answer: E.

Hope it's clear.


Sorry for being silly but are we asked probability or total number of combinations ?


Does it change the solution in any way?
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Re: Company X is forming a two-person committee by selecting one  [#permalink]

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New post 16 Oct 2014, 12:29
diebeatsthegmat wrote:
to answer this question, the big question is how many managers are there in the company X and the form: P(m)+P(b)-P(m&b)-None
statement 1 lets you know none is 30 surely insufficient
statement 2 lets you know both P(m&b)= 10 insufficient
both cant give you the answer of the question how many manahers there are in the company X
so E


slight typo i think. Neither is added back in sets, isn't it? Technically, even in probability it should be added back. Can someone correct me please?
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Company X is forming a two-person committee by selecting one  [#permalink]

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New post 04 Mar 2017, 08:48
stunn3r wrote:
martie11 wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.


I know everyone got it right but there is more to this question .. this is a good question ..

here:
no. of managers - M
Board members - B

it can be represented in a venn Diagram ...
a+c = M(a - only managers, c - managers as well as board member)
b+c = B)b - only board members, c - manafers as well as board members)

statement1:
aC1.bC2 = 30(selected one manager and one board member)

a.b = 60 .. now there are many combinations for a.b such as 1*60, 2*30, 3*20 ...... hence a+b can be 61,32,23 ..... not sufficient

statement2:
cC2=10 (selected 2 persons which are both manager and board member)
c(c-1) = 20 .. only possible answer for c is 5 ..



... If the question would've asked no. of members which are both manager and board member we could've answered it with B.



can you please explain how is cC2=10 translated into c(c-1) = 20 ?
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Re: Company X is forming a two-person committee by selecting one  [#permalink]

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New post 05 Mar 2017, 02:27
HassanGmat17 wrote:
stunn3r wrote:
martie11 wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?

(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.


I know everyone got it right but there is more to this question .. this is a good question ..

here:
no. of managers - M
Board members - B

it can be represented in a venn Diagram ...
a+c = M(a - only managers, c - managers as well as board member)
b+c = B)b - only board members, c - manafers as well as board members)

statement1:
aC1.bC2 = 30(selected one manager and one board member)

a.b = 60 .. now there are many combinations for a.b such as 1*60, 2*30, 3*20 ...... hence a+b can be 61,32,23 ..... not sufficient

statement2:
cC2=10 (selected 2 persons which are both manager and board member)
c(c-1) = 20 .. only possible answer for c is 5 ..



... If the question would've asked no. of members which are both manager and board member we could've answered it with B.



can you please explain how is cC2=10 translated into c(c-1) = 20 ?


\(C^2_c=\frac{c!}{(c-2)!*2!}=\frac{(c-2)!*(c-1)*c}{(c-2)!*2!}=\frac{(c-1)*c}{2!}\)

\(\frac{(c-1)*c}{2!}=10\) --> \((c-1)*c=20\).

Hope it's clear.
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Re: Company X is forming a two-person committee by selecting one  [#permalink]

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New post 24 Nov 2017, 11:23
Let a - number of persons who are only managers
b - number of persons who are both manager and board member
c - number of persons who are only board member

Question: N(atleast one committee member is both a manager and a board member) = N(both are both manager and board members) + N(one member is both manager and board memeber AND other member is only manager) +
N(one member is both manager and board memeber AND other member is only board member)

= \(bC2 + bC1 * aC1 + bC1 * cC1 = bC2 + b * a + b * c = bC2 + b * (a+c)\) = ?

Statement 1:
The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
= number of persons only manager * number of persons only board member = 30
= \(aC1 * cC1 = 30\) => \(ac = 30\), we don't know about b -> Insuff

Statement 2:

The number of committees that can be formed such that each committee member is both a manager and a board member is 10.
= bC2 = 10, we can find b = 5, we don't know about a, c InSuff

Statement 1 + 2:
\(bC2 + b * (a+c) = 10 + 5 (a + c)\)
\(ac = 30, => a = 15, c = 2, OR a = 6, c = 5\)
accordingly we get different values of \(10 + 5 (a + c)\)
Still not sufficient

Answer (E)
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