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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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"

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

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.

E should be the OA since we do not know the number of managers and board members

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.

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.

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

You are GREAT !

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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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?

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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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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