Counting Permutations for Board of Directors: 10 Members and 5 Officers

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Homework Help Overview

The discussion revolves around counting permutations for a board of directors consisting of 10 members, from which 5 officers are to be elected, including specific roles such as president, vice-president, secretary, and treasurer. The problem includes variations based on the presence of physicians among the members.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of combinations for different scenarios, including the total number of officer combinations and the implications of having physicians in specific roles. There is a focus on the importance of order and the uniqueness of combinations based on who holds each position.

Discussion Status

Some participants have offered guidance on rethinking the approaches for specific parts of the problem, particularly regarding the counting of combinations when a physician is assigned to a role. There is an ongoing exploration of how to account for different combinations and the implications of the roles assigned to physicians.

Contextual Notes

Participants are considering the implications of the roles assigned to physicians and how that affects the counting of combinations. There is also mention of the need to clarify whether certain combinations are considered unique based on the specific individuals assigned to each role.

chris2020
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Homework Statement


Board of directors has 10 members. from the 10 members they will elect 5 officers. President, vice-pres, sec and treas

A) From the 10 board members how many combinations of officers are there?

B) If three board memebers are physicians, how many combinations have a physician as president?

C) Exactly 1 physician as an officer?

D) At least 1 physician as an officier?

Homework Equations

m[/B]
M×N

The Attempt at a Solution


A) 10×9×8×7×6, 10!/5!

B) 3x9x8x7x6

C) 3×7×6×5×4

D) 3×9×8×7×6

I just imagined plugging in people to the various office positions and used the simple M×N formula.
 
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Your approach is fine - for A you reduced the number each time because nobody can have two positions, and divided by 5! because the order is unimportant.

For B: While there are three ways a physician can be president, you don't care which physician is president. Remember also that the order they get assigned still does not matter.
 
So ur saying i should rethink solutions for b, c, d?
 
It can help to check using letters and numbers:

ABCDEFGHIJ are the members
12345 are the jobs. 1 = president.

ABC are doctors X = placeholder

ADEFG
BDEFG
CDEFG
... each have a physician as president ... do we care if it's A or B or C who holds the post when it comes to counting unique combinations.
I think that's a judgement call - also is ADEFG the same or different from ADFEG?
 
Simon Bridge said:
do we care if it's A or B or C who holds the post when it comes to counting unique combinations.
Yes, they count as different combinations here. I agree with the OP's answer for (b).
The answer for (c) is not quite right. Chris, can you explian your reasoning there?
The answer to (d) should be much higher than the answer to (b).
 
It seems any good countable combination must have A, B or C in the president position however the other positions can permute all combinations a-j... well minus whomever is president in that combination...c)

I was thinking i have 2 pools of people. Physicians (3) and non-physicians(7)
so i just plugged a physician in an office then used the 7 from the other pool to permute the other offices..
 
chris2020 said:
It seems any good countable combination must have A, B or C in the president position however the other positions can permute all combinations a-j... well minus whomever is president in that combination...c)

I was thinking i have 2 pools of people. Physicians (3) and non-physicians(7)
so i just plugged a physician in an office then used the 7 from the other pool to permute the other offices..

For c) you have 3 different physicians and 5 different offices to fill in the one physician. Then populate the rest of the offices. For d) think about how many ways of populating use no physicians. The rest of them use at least one.
 

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