how many committees? (counting questions...)

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June 28th 2011, 05:29 PM
iamthemanyes

how many committees? (counting questions...)

From a group of 11 people (5 men and 6 women), we have to form 3
committees. Committee A and B consist of 4 people and committee C consists
of 3 people. How many diﬀerent possibilities are there
a) in total.
b) if one of the men does not want to be in the committee A.
c) if one of the men does not want to be in the committee A and one of the
women does not want to be in the committee C.
d) if two of the women do not want to be in the same committee.
e) if one of the men and one of the women do not want to be in the same
committee.
f ) if committee C has to consist only of women.
g) if there has to be at least one men in each committee.

(any form of help is appreciated. thanks.)
June 28th 2011, 07:13 PM
Soroban

Re: how many committees? (counting questions...)

Hello, iamthemanyes!

I'll get you started . . .

Quote:

From a group of 11 people (5 men and 6 women), we will form 3 committees.
Committee A and B have 4 people each, and committee C has 3 people.
How many diﬀerent possibilities are there

a) in total.

$\text{There are: }\:{11\choose4,4,3} \:=\:\frac{11!}{4!\,4!\,3!} \:=\:11,\!550\text{ possibilities.}$

Quote:

b) if one of the men does not want to be in the committee A.

$\text{Suppose his name is }\times.$

Place him on committee $B.$

We have: . $\begin{array}{|c|c|c|} A & B & C \\ ---- & \times--- & --- \end{array}$
The other 10 people can be assigned in: . ${10\choose4,3,3} \:=\:4200$ ways.

Place him on committee $C.$

We have: . $\begin{array}{|c|c|c|} A & B & C \\ ---- & ---- & \times-- \end{array}$
The other 10 people can be assigned in: . ${10\choose4,4,2} \:=\:3150$ ways.

$\text{Therefore, there are: }\:4200 + 3150 \:=\:7350\text{ ways}$
. . $\text{in which }\times\text{ is }not\text{ on committee A.}$
June 29th 2011, 06:08 AM
Soroban

Re: how many committees? (counting questions...)

Hello again, iamthemanyes!

Here are a couple more . . .

Quote:

From a group of 5 men and 6 women, we form 3 committees.
Committee A and B have 4 people each, and committee C has 3 people.

How many diﬀerent possibilities are there

d) if two of the women do not want to be in the same committee.

$\text{There are }11,\!550\text{ possible committees.}$

$\text{Call the women }a\text{ and }b.$
$\text{In how many ways can they be on the }same\text{ committee?}$

$\text{(1) Place them on committee A.}$

We have: . $\begin{array}{|c|c|c|} A & B & C \\ ab-- & ---- & --- \end{array}$
$\text{The other 9 people can be assigned in: }\:{9\choose2,4,3} \:=\:1260\text{ ways.}$

$\text{(2) Place them on committee B.}$

We have: . $\begin{array}{|c|c|c|} A & B & C \\ ---- & ab-- & --- \end{array}$
$\text{The other 9 people can be assigned in: }\:{9\choose4,2,3} \:=\:1260\text{ ways.}$

$\text{(3) Place them on committee C}$

We have: . $\begin{array}{|c|c|c|} A & B & C \\ ---- & ---- & ab- \end{array}$
$\text{The other 9 people can be assigned in: }\:{9\choose4,4,1} \:=\:630\text{ ways.}$

$\text{Hence: }\:1260 + 1260 + 630 \:=\:3150\text{ ways that }a\text{ and }b\text{ are together.}$

$\text{Therefore: }\:11,\!550 - 3150 \:=\:8400\text{ ways that }a\text{ and }b\text{ are }not\text{ together.}$

Quote:

f) if committee C has to consist only of women.

Select 3 women to be on committee C: . ${6\choose3} \,=\,20\text{ choices.}$
We have: . $\begin{array}{|c|c|c|} A & B & C \\ ---- & ---- & www \end{array}$
$\text{The other 8 people can be assigned in: }\:{8\choose4,4} \:=\:70\text{ ways.}$

Therefore: . $20\cdot 70 \:=\:1400\text{ ways.}$

Quote:

g) if there has to be at least one man in each committee.

Consider the opposite situation: there is a committee that is all women.

Suppose committee A is all women.
Choose 4 women to be on committee A: . ${6\choose4} \:=\:15\text{ choices}$
We have: . $\begin{array}{|c|c|c|} A & B & C \\ wwww & ---- & --- \end{array}$
$\text{The other 7 people can be assigned in: }\:{7\choose4,3} \:=\:35\text{ ways.}$
. . $\text{Hence: }\:15\cdot 35 \:=\:525\text{ ways.}$

Suppose committee B is all women.
Choose 4 women to be on committee B: . ${6\choose4} \:=\:15\text{ choices}$
We have: . $\begin{array}{|c|c|c|} A & B & C \\ ---- & wwww & --- \end{array}$
$\text{The other 7 people can be assigned in: }\:{7\choose4,3} \:=\:35\text{ ways.}$
. . $\text{Hence: }\:15\cdot 35 \:=\:525\text{ ways.}$

Suppose committee C is all women.
We already solved this in part (f): $1400\text{ ways.}$

$\text{Hence: }\:525 + 525 + 1400 \:=\:2450\text{ ways to have a committee with }no\text{ men.}$

$\text{Therefore: }\:11,\!550 - 2450 \:=\:9100\text{ ways}$
. . $\text{to have at least one man on each committee.}$