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Anthony and Michael sit on the six-member board of directors [#permalink]
 08 Dec 2006, 09:16
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Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
A)20%
B)30%
C)40%
D)50%
E)60%
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possible subcommittees that include Michael = 6C2
possible subcommittees that include Michael also include Anthony= 6C1
(6C1/ 6C2) * 100 = 40%
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Senior Manager
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AK wrote: possible subcommittees that include Michael = 6C2
possible subcommittees that include Michael also include Anthony= 6C1
(6C1/ 6C2) * 100 = 40%
Could you plz explain more. I am having difficulties following it. thanks
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Re: DS - board of directors [#permalink]
08 Dec 2006, 16:27
Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
(1st case)
M and A both are already member means we need to find commitees with 1 person of rest of 4
4C1 i.ie 4 ways to form a comitee.
2 comitees i.e 8 ways
(2nd case)
if michal is included already
5 people with 2 people comitee = 5C2=5*4/2=10
10 ways to form 1 comitte
togather 20 ways for 2 comittes
8/20= 2/5=40%
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Senior Manager
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arjsingh1976 wrote: AK wrote: possible subcommittees that include Michael = 6C2
possible subcommittees that include Michael also include Anthony= 6C1
(6C1/ 6C2) * 100 = 40% Could you plz explain more. I am having difficulties following it. thanks Damager has explained well.. I did a mistake while puting it here
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Subcommittees that include Michael = 5C2 = 10 (select 2 from 5)
Subcommittees that include both Michael and Anthony= 4C1 = 4 (select 1 from 4; )
Answer = (4/10)*100 = 40%
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I am really confused by this ! It is not making sense
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I am really confused
I am usually good with combo problems
How do we get to these ?
Subcommittees that include Michael = 5C2 = 10 (select 2 from 5)
Quote: Subcommittees that include both Michael and Anthony= 4C1 = 4 (select 1 from 4; )
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how abt this...
one sub committee has 3 ppl nd u need to both anthony nd michael on it....
So..
3C2/6C2 or(+) 3C2/6C2...giving us 2/5 nd thereby 40%
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jamesrwright3 wrote: I am really confused I am usually good with combo problems How do we get to these ? Subcommittees that include Michael = 5C2 = 10 (select 2 from 5) Quote: Subcommittees that include both Michael and Anthony= 4C1 = 4 (select 1 from 4; ) Comitee size =3
Michael has to be included. So he has already been chosen, i seat taken.
we need to find number of ways to choose members for remaing 2 seats.
Total number of people 6 - michel ( as he is already chosen = 5
Number of ways to choose 2 person out of 5 = 5C2= 5!/3!2!=5*4*3*2/( 3*2*2)===> 10
Similarly
When both michal and antony are chosen
total seats 3 . antony takes 1 snd micheal takes 1 . 1 seat remaining
Out of 6 people antony and michel alreasy chosen reming people 4
Number of ways to choose 1 peron out of 4 is 4C1=4!/3!=4
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Damager wrote: jamesrwright3 wrote: I am really confused I am usually good with combo problems How do we get to these ? Subcommittees that include Michael = 5C2 = 10 (select 2 from 5) Quote: Subcommittees that include both Michael and Anthony= 4C1 = 4 (select 1 from 4; ) Comitee size =3 Michael has to be included. So he has already been chosen, i seat taken. we need to find number of ways to choose members for remaing 2 seats. Total number of people 6 - michel ( as he is already chosen = 5 Number of ways to choose 2 person out of 5 = 5C2= 5!/3!2!=5*4*3*2/( 3*2*2)===> 10 Similarly When both michal and antony are chosen total seats 3 . antony takes 1 snd micheal takes 1 . 1 seat remaining Out of 6 people antony and michel alreasy chosen reming people 4 Number of ways to choose 1 peron out of 4 is 4C1=4!/3!=4 Thanks. I got it now!
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arjsingh1976 wrote: Damager wrote: jamesrwright3 wrote: I am really confused I am usually good with combo problems How do we get to these ? Subcommittees that include Michael = 5C2 = 10 (select 2 from 5) Quote: Subcommittees that include both Michael and Anthony= 4C1 = 4 (select 1 from 4; ) Comitee size =3 Michael has to be included. So he has already been chosen, i seat taken. we need to find number of ways to choose members for remaing 2 seats. Total number of people 6 - michel ( as he is already chosen = 5 Number of ways to choose 2 person out of 5 = 5C2= 5!/3!2!=5*4*3*2/( 3*2*2)===> 10 Similarly When both michal and antony are chosen total seats 3 . antony takes 1 snd micheal takes 1 . 1 seat remaining Out of 6 people antony and michel alreasy chosen reming people 4 Number of ways to choose 1 peron out of 4 is 4C1=4!/3!=4 Thanks. I got it now! Awesome explanation
Too bad we can't give people reputation on this board:)
Thanks!
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Re: DS - board of directors [#permalink]
10 Dec 2006, 17:57
arjsingh1976 wrote: Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
A)20%
B)30%
C)40%
D)50%
E)60%
Is the answer 40% ? I am getting 20%. Well from what i understand it is
(No. Subcommittess including both Michael and Anthony) / (No. of all possible Subcommittees)
Which is 4C1 / 6C3 = 4/20 = 20%
What you guys are talking about is No. of subcommitees which has Michael, and not all possible subcommitees. So that number is 10
and that way u get 40%.
Please explain !
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Re: DS - board of directors [#permalink]
10 Dec 2006, 18:06
arjsingh1976 wrote: Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
Your equation should be:
subcomittees that include Michael+Anthony/no. with Michael + 2 others
Last edited by MBAlad on 10 Dec 2006, 18:25, edited 1 time in total.
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Re: DS - board of directors [#permalink]
10 Dec 2006, 18:15
das_go wrote: arjsingh1976 wrote: Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
A)20%
B)30%
C)40%
D)50%
E)60% Is the answer 40% ? I am getting 20%. Well from what i understand it is (No. Subcommittess including both Michael and Anthony) / (No. of all possible Subcommittees) Which is 4C1 / 6C3 = 4/20 = 20% What you guys are talking about is No. of subcommitees which has Michael, and not all possible subcommitees. So that number is 10 and that way u get 40%. Please explain ! I solved this problem the same way as you did. However, I think the number of all possible subcommittees would be 10 instead of 20. (6C3)/2.
You are correct on the number of all possible sub committee if out of 6, you only have to pick 3 people. However, when the 6 people are SPLIT into 2 subgroupd, then when you pick out the 3 people, the other 3 left would also set one arrangement. so to avoid double counting, we need to divide that amount by 2. So, 20/2 = 10. the rest is the same.
Hope that makes sense.
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All possible with Michael = 5C2 = 5*4/2*1 = 10
As you've said, by selecting one individual subcommittee you leave a corresponding unique subcommittee so total number of unique subcommittees = 10
Note: We use 5C2 (not 6C3) because we are looking at total number of subcommittees with Michael, not the total number possible. Therefore, Michael has already taken the first seat so we look at the number of ways to choose two people from the remaining 5.
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