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# Six doctors, six nurses and six anesthesiologists are at a

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Six doctors, six nurses and six anesthesiologists are at a [#permalink]

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30 Sep 2005, 19:45
00:00

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(N/A)

Question Stats:

100% (01:00) correct 0% (00:00) wrong based on 1 sessions

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Six doctors, six nurses and six anesthesiologists are at a conference. In one of the sessions the group must form teams of two doctors, two nurses and an anesthesiologist. How many such committees can be formed if one of the nurses has a restaining order against one of the doctors and therefore refuses to work with him?

(A) 2400
(B) 1800
(C) 1200
(D) 600
(E) 216
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01 Oct 2005, 00:43
c) 1200

6C2*6C2*6 - 5*5*6 = 1200

groups of 2 doctors, 2 nurses, and an anesthesiologist, minus the groups which have the restrained doctor and nurse (5*5*6)
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01 Oct 2005, 05:46
Thanks chet. It was that last part - 5*5*6 that was the most confusing.

OA is C.
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03 Oct 2005, 09:41
chets wrote:
c) 1200

6C2*6C2*6 - 5*5*6 = 1200

groups of 2 doctors, 2 nurses, and an anesthesiologist, minus the groups which have the restrained doctor and nurse (5*5*6)

can you explain 5*5*6 part please. I couldnt figure out and tries this POE.

i picked all possible combinations sans any condition:
6c2 * 6c2 * 6c1 = 1350. The answer will be less than this.
this eliminates all options except :
C and remotly D.

find all combination excluding problem nurse:
6c1 * 5c1 * 6c1 = 900

Eliminate D.
ans: C
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03 Oct 2005, 10:12
The 5*5*6 part identifies 3 groups. And focuses on the execptions of the problem. 1 nurse can't work with 1 doctor.

1 doctor chosen(bad doctor)* 5 other choices* 1 nurse chosen(helpless victim)* 5 other choices* 6 possible aneth.
1*5*1*5*6=5*5*6=150 choices removed.

Hope this clears it up.

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03 Oct 2005, 11:26
jrabenho wrote:
The 5*5*6 part identifies 3 groups. And focuses on the execptions of the problem. 1 nurse can't work with 1 doctor.

1 doctor chosen(bad doctor)* 5 other choices* 1 nurse chosen(helpless victim)* 5 other choices* 6 possible aneth.
1*5*1*5*6=5*5*6=150 choices removed.

Hope this clears it up.

got it. thanks.

find out all combinations where problem doc and helpless nurse together:
Dont count them. This leaves 5 docs, 5 nurses, 6 anesthe or 5 * 5 * 6 combinations.
subtract this to get all combinations where they are not together
03 Oct 2005, 11:26
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