My Math Forum choosing question

 March 22nd, 2010, 07:02 PM #1 Senior Member   Joined: Sep 2009 Posts: 115 Thanks: 0 choosing question A student council consists of three freshmen, four sophomores, three juniors, and five seniors. How many committees of eight members can be formed containing at least one member from each class? so we know the total number of ways is : 16 choose 8. Then I am not sure if this is right but do we find the number that contain none from each class such as: none from freshmen would be 13 choose 8 none from sophomores would be 12 choose 8 none from juniors would be 13 choose 8 none from seniors would be 11 choose 8 Hence, we subtract now: [16 choose 8] - ([13 choose 8] + [12 choose 8] + [13 choose 8] + [11 choose 8]) I am not sure if this is right. Thank you in advance.
 March 22nd, 2010, 07:45 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: choosing question It's not quite right, because you've subtracted off (for example) councils with only freshmen and sophomores twice.
March 23rd, 2010, 09:09 AM   #3
Math Team

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Re: choosing question

Hello, wannabe1!

This is more complicated than you or I thought.

Quote:
 A student council consists of 3 freshmen, 4 sophomores, 3 juniors, and 5 seniors. How many committees of eight members can be formed containing at least one member from each class?

$\text{I will call the classes: }\:A,\,B,\,C,\,D.$

$\text{We have: }\;\begin{array}{cc} A& 3 \\ \\ \\ B & 4 \\ \\ \\ C & 3 \\ \\ \\ D & 5 \\ \\ \\ \hline \\ \\ \text{Total} & 15 \end{array}$

$\text{There are }15\text{ members: }\:{15\choose8}\text{ possible committees.}$

$\text{We can count these four possible scenarios:}$

[color=beige]. . [/color]$\begin{array}{cc}\text{(1) no }A's: &{12\choose8}\text{ ways} \\ \\ \\ \text{(2) no }B's: &{11\choose8}\text{ ways} \\ \\ \\ \text{(3) no }C's: & {12\choose8}\text{ ways} \\ \\ \\ \text{(4) no }D's: & {10\choose8}\text{ ways} \end{array}$

$\text{But these four cases have some "overlap".$
$\text{We must count and eliminate these duplications.}$

$\text{No A's and no B's}$
[color=beige]. . [/color]$\text{There are still 3 C's and 5 D's: }\;\text{ 1 possible committee}$

$\text{No A's and no C's}$
[color=beige]. . [/color]$\text{There are still 4 B's and 5 D's: }\:{9\choose 8}\,=\,9\text{ possible committees}$

$\text{No A's and no D's}$
[color=beige]. . [/color]$\text{There are only 4 B's and 3 C's: }\:\text{ no committees possible}$

$\text{No B's and no C's}$
[color=beige]. . [/color]$\text{There are still 3 A's and 5 D's: }\;\text{ 1 possible committee}$

$\text{No B's and no D's}$
[color=beige]. . [/color]$\text{There are only 3 A's and 3 C's: }\:\text{ no committees possible}$

$\text{No C's and no D's}$
[color=beige]. . [/color]$\text{There are only 3 A's and 4 B's: }\:\text{ no committees possible}$

$\text{Hence, there are: }\:1 \,+\, 9 \,+\, 1 \:=\:11\text{ in the "overlap=".}=$

$\text{The answer is: }\;{15\choose8} \,-\,{12\choose8} \,-\,{11\choose8}\,-\,{12\choose8}\,-\,{10\choose8}\,+\,11$

Well, I hope my reasoning is correct!
[color=beige] .[/color]

 March 23rd, 2010, 09:55 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: choosing question Generally, you;d have to continue with the number of councils with no Fs, Ss, or Js; the councils with no Fs, Ss, or Srs; the councils with no Fs, Js, or Srs; the councils with no Ss, Js, or Srs. But in this case there are none, so the answer looks good to me. Lookup "inclusion-exclusion" for more.
 March 23rd, 2010, 04:24 PM #5 Senior Member   Joined: Sep 2009 Posts: 115 Thanks: 0 Re: choosing question o ya your right, thanks that is really helpful (ps. I was doing it with 4 juniors that is why i chose from 16 members) and you are correct about the overlap I totally forgot. Thank you again.

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