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A certain board has 12 persons: the president, two

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A certain board has 12 persons: the president, two [#permalink] New post 05 Mar 2003, 23:58
A certain board has 12 persons: the president, two vice-presidents, eight managers, and state inspector. A comission of 6 people should be taken. Conditions: two vice-presidents have both to be taken, the inspector and the president cannot be taken together. How many ways are there to do so?

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 [#permalink] New post 06 Mar 2003, 06:58
is the answer 182

2*8c3+8c4
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 [#permalink] New post 06 Mar 2003, 07:04
looks OK

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 [#permalink] New post 19 Jun 2003, 21:22
OR

All combinations - forbidden ones

2 VP are a must, so we have the following matrix VV****, or 10C4 ways to fill it. P and S are not together, so FORBIDDEN= VVPS**, or 8C2 ways tp fill it.

Finally, 10C4–8C4=210–28=182
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 [#permalink] New post 08 Feb 2006, 09:54
I am lost, can someone please explain the solution. Thanks

Last edited by Riuscita on 08 Feb 2006, 10:14, edited 1 time in total.
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 [#permalink] New post 08 Feb 2006, 13:40
ashish_madan wrote:
is the answer 182

2*8c3+8c4


What is the addition of 8c4 for?

6 places, 2 VP already fixed, if secretary is present , president cannot be there and vice-versa. Remaining 3 places have to be filled with 8 manager available i.e. 8c3

For both cases it will be 2*8c3

Although i should admit that i do not found anything wrong in Stoylar approach.

What is the mistake i am doing?
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Re: combi 4 [#permalink] New post 08 Feb 2006, 18:10
stolyar wrote:
A certain board has 12 persons: the president, two vice-presidents, eight managers, and state inspector. A comission of 6 people should be taken. Conditions: two vice-presidents have both to be taken, the inspector and the president cannot be taken together. How many ways are there to do so?
:panel

like it.
p=1
vp=2
m=8
i=1

= (8c3) + (8c3) + 8c4
=2 (8c3) + 70= 2x56+70=182

Last edited by Professor on 08 Feb 2006, 18:18, edited 1 time in total.
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 [#permalink] New post 08 Feb 2006, 18:13
prash_c wrote:
ashish_madan wrote:
is the answer 182

2*8c3+8c4


What is the addition of 8c4 for?


total required = 6, which can be taken in three ways.

1. p + 2 vps + 3 out of 8 managers, not 9 people because you cannot include inspector. so exclude him

2. Inspector + 2 vps + 3 out of 8 managers. now exclude p.

3. 2 vps + 4 out of 8 managers. now exclude p and inspector.

add all these combinations. yours is third case.
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 [#permalink] New post 09 Feb 2006, 00:52
Looking at previous posts seems that my logic is wrong but still i will give it a try.
2 VP's are included so we need 4 more people. But another restriction is that SI and P can not be taken together. Then the other 4 members will have to be selected out of 9, which can be done in 126 ways=9C4. But this result should be multiplied by 2 since one time SI is excluded, other P is excluded from the group of 10
So the final outcome comes to be 252
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 [#permalink] New post 09 Feb 2006, 01:08
BG wrote:
Looking at previous posts seems that my logic is wrong but still i will give it a try.
2 VP's are included so we need 4 more people. But another restriction is that SI and P can not be taken together. Then the other 4 members will have to be selected out of 9, which can be done in 126 ways=9C4. But this result should be multiplied by 2 since one time SI is excluded, other P is excluded from the group of 10
So the final outcome comes to be 252


I believe 2 vice presidents is always a must, therefore in one case it is 1) VP VP I x x x (but cannot be P) in other case 2) VP VP P x x x (but cannot be I).
All in all we have 12 persons.
In the first case we have 9 people left to be selected to 3 places. However, we cannot choose President in that place. Therefore, 8 persons are left to be selected in 3 places.
Similarly is the case for the second case, where 9 people are left to be selected in 3 places, however the inspector cannot be selected. We are left with 8 persons to 3 places.

Now 8c4 is the situation when in this group of 6 members we do not have President nor the inspector. Therefore, VP VP x x x x which initially says that there are 4 places and 10 people to be selected, when we throw out :) president and inspector we are left with 8 people to be selected to 4 places. Here we have a formula 8c4.

So 8c3+8c3+8c4 is the answer.......
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 [#permalink] New post 09 Feb 2006, 03:52
Thanks for your post SimaQ, but the stem says that
"two vice-presidents have both to be taken, the inspector and the president cannot be taken together. How many ways are there to do so"
which as far as i understand it , means that it is not necessary that either of them should participate. That is why we are left with 9 options, to select 4 people.
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 [#permalink] New post 09 Feb 2006, 04:55
BG wrote:
Thanks for your post SimaQ, but the stem says that
"two vice-presidents have both to be taken, the inspector and the president cannot be taken together. How many ways are there to do so"
which as far as i understand it , means that it is not necessary that either of them should participate. That is why we are left with 9 options, to select 4 people.


Yes, it is not necessary that either president or inspector participates, but it is only one (out of three possibilities). In your previous post you consider that either president or inspector participates... You do not assume that either of them does not participate....

+ it is not correct when you say that you choose 9 people into 4 places, because.... 12 - 2VP=10 people. If you choose President for another place you are left with 10-1=9 people, but you cannot have inspector in this group then 9-1=8 people. Now, the 3 places are occupied and the 3 are left, so 8c3...

If you say 9c4 you say that 2VP are present 1P/I is present and that you can choose any 4 persons (which is wrong because the total number of places is 6 and 3 places are already occupied) to the rest of the board (which is not correct because you cannot choose I if P is present and vice versa)
  [#permalink] 09 Feb 2006, 04:55
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