How Many Ways Can Club Officers Be Selected Under Various Conditions?

[tangodirt]

Homework Statement

A president, treasurer, and secretary, all different, are to be chosen from among the 10 active members of a university club. How many different choices are possible if:

a) There are no restrictions.
b) A will serve only if she is the treasurer.
c) B and C will not serve together.
d) D and E will serve together or not at all.
e) F must be an officer.

Homework Equations

N/A

The Attempt at a Solution

For part (a): 720 combinations.
For part (b): 72 combinations.
For part (c): 448 combinations.
For part (d): 8 combinations.
For part (e): 72 combinations.

Are any of these answers correct? I am wary about the answer for part (d), but please let me know which ones might be wrong and I will explain my methodology. Thanks!

 

[HallsofIvy]

I get "720" for (a) but different answers from you for all others.

 

[tangodirt]

Okay, I have a feeling I know where I've gone wrong. Before I continue further, I reworked part (b) and got 576 permutations. Does that sound right? Here's my work:

Assumptions:
- order counts
- no replacements

Condition 1: if A is elected treasurer:

$p = \frac{9!}{(9-2)!}$

p = 72 permutations

Condition 2: if A is not elected treasurer:

$p = \frac{9!}{(9-3)!}$

p = 504 permutations

Total Permutations:

$p_{total} = 72+504$

p_total = 576 permutations

 

[Dick]

Okay, I have a feeling I know where I've gone wrong. Before I continue further, I reworked part (b) and got 576 permutations. Does that sound right? Here's my work:

Assumptions:
- order counts
- no replacements

Condition 1: if A is elected treasurer:

$p = \frac{9!}{(9-2)!}$

p = 72 permutations

Condition 2: if A is not elected treasurer:

$p = \frac{9!}{(9-3)!}$

p = 504 permutations

Total Permutations:

$p_{total} = 72+504$

p_total = 576 permutations

That looks ok to me for b).

 

[tangodirt]

So, after reworking the problems, here is what I am getting:

(a) = 720
(b) = 576
(c) = 480
(d) = 344
(e) = 72

The only one that remained the same was part (e). Here's how I solved it:

Assumptions:
- order counts
- no replacements

$p = \frac{9!}{(9-2)!}$

p = 72 permutations

Any thoughts on my results?

 

[Dick]

So, after reworking the problems, here is what I am getting:

(a) = 720
(b) = 576
(c) = 480
(d) = 344
(e) = 72

The only one that remained the same was part (e). Here's how I solved it:

Assumptions:
- order counts
- no replacements

$p = \frac{9!}{(9-2)!}$

p = 72 permutations

Any thoughts on my results?

Sorry, but I'm only checking the ones where you gave some idea how you did it. But for that one there are three different ways F can be a officer, right? F could be president, treasurer, or secretary.

 

[tangodirt]

Sorry, but I'm only checking the ones where you gave some idea how you did it. But for that one there are three different ways F can be a officer, right? F could be president, treasurer, or secretary.

Good call.

Here's the other solutions.

--------------------------------------------------
Part (a):
$p = \frac{10!}{(10-3)!}$

p = 720 permutations
--------------------------------------------------

Part (b):
if A is treasurer:
$p = \frac{9!}{(9-2)!}$

p = 72 permutations

if A is not treasurer:
$p = \frac{9!}{(9-3)!}$

p = 504 permutations

total permutations:
p = 72 + 504 = 576 permutations
--------------------------------------------------

Part (c):
if B is an officer:
$p = \frac{9!}{(9-2)!}$

p = 72 permutations

if C is an officer:
$p = \frac{9!}{(9-2)!}$

p = 72 permutations

if neither are officers:
$p = \frac{8!}{(8-3)!}$

p = 336 permutations

total:
p = 72 + 72 + 336 = 480 permutations
--------------------------------------------------

Part (d):
if D and E are officers:
$p = \frac{8!}{(8-1)!}$

p = 8 permutations

if D and E are not serving:
$p = \frac{8!}{(8-3)!}$

p = 336 permutations

total:
p = 336 + 8 = 344 permutations
--------------------------------------------------

Part (e):

$p = \frac{9!}{(9-2)!}$

p = 72 permutations / position * 3 positions = 216 permutations

 

[HallsofIvy]

I am concerned that you appear automatically to be using the permutations formula. Here's how I would do (b). If A is treasurer, then there are 9 choices for president and then 8 choices for secretary. There are 9(8)= 72 ways to do that. If A is not treasurer, then there are 9 choices (everyone except A) for president, then 8 choices for treasurer, then 7 choices for secretary. There are 9(8)(7)= 504 ways to do that. Together there are 72+ 504= 576 ways to choose officers if A must be treasurer or not an officer.

 

[Dick]

c) and d) have they same sort of problem e) used to have. If someone is going to serve as an officer there are three different ways to do that. You should also check the counts of people available to fill the officers positions.

 

[tangodirt]

I am concerned that you appear automatically to be using the permutations formula. Here's how I would do (b). If A is treasurer, then there are 9 choices for president and then 8 choices for secretary. There are 9(8)= 72 ways to do that. If A is not treasurer, then there are 9 choices (everyone except A) for president, then 8 choices for treasurer, then 7 choices for secretary. There are 9(8)(7)= 504 ways to do that. Together there are 72+ 504= 576 ways to choose officers if A must be treasurer or not an officer.

Well, I'm glad I get the same answer as you. I like thinking about the problem in the same manner as you did, but my teacher prefers the "mathematical" solution. It makes sense, though too:

$p = \frac{(sample size)!}{(sample size - open positions)!}$

 

[tangodirt]

c) and d) have they same sort of problem e) used to have. If someone is going to serve as an officer there are three different ways to do that. You should also check the counts of people available to fill the officers positions.

Good catch. Thanks for pointing that out. The only issue now is that my total number of permutations under this scenario exceeds the total number of permutations without restrictions:

If B is an officer and C is not, then the sample size is reduced by one and the number of open positions is also reduced by one (B takes a position):

$p = \frac{9!}{(9-2)!}$

p = 72 permutations per officer position = 216 permutations

Since B and C are interchangeable, the number of permutations for C serving and B not is identical.

Then, for the case the B and C both do not serve, then the sample size is reduced by two and the number of open positions is increased to three:

$p = \frac{8!}{(8-3)!}$

p = 336 permutations

Therefore, the total number of permutations is the sum of all other permutations, so that:

p_total = 216 + 216 + 336 = 912 permutations.??

 

[Dick]

Good catch. Thanks for pointing that out. The only issue now is that my total number of permutations under this scenario exceeds the total number of permutations without restrictions:

If B is an officer and C is not, then the sample size is reduced by one and the number of open positions is also reduced by one (B takes a position):

$p = \frac{9!}{(9-2)!}$

p = 72 permutations per officer position = 216 permutations

Since B and C are interchangeable, the number of permutations for C serving and B not is identical.

Then, for the case the B and C both do not serve, then the sample size is reduced by two and the number of open positions is increased to three:

$p = \frac{8!}{(8-3)!}$

p = 336 permutations

Therefore, the total number of permutations is the sum of all other permutations, so that:

p_total = 216 + 216 + 336 = 912 permutations.??

Good observation. But if B is an officer and C is not, do you really have 9 people left to fill the remaining offices?