Six doctors, six nurses and six anesthesiologists are at a

Oldest

Best Reply

Author

Message

TAGS:

Current Student

Joined: 29 Jan 2005

Posts: 5289

Followers: 17

Kudos [?]: 90 [0], given: 0

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

30 Sep 2005, 19:45

00:00

Question Stats:

0% (00:00) correct

0% (00:00) wrong

based on 0 sessions

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

Manager

Joined: 15 Jul 2005

Posts: 106

Followers: 1

Kudos [?]: 1 [0], given: 0

[#permalink]

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)

Current Student

Joined: 29 Jan 2005

Posts: 5289

Followers: 17

Kudos [?]: 90 [0], given: 0

[#permalink]

01 Oct 2005, 05:46

Thanks chet. It was that last part - 5*5*6 that was the most confusing.

OA is C.

Joined: 22 Aug 2005

Posts: 1133

Location: CA

Followers: 1

Kudos [?]: 9 [0], given: 0

[#permalink]

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

Intern

Joined: 08 Feb 2005

Posts: 15

Followers: 0

Kudos [?]: 0 [0], given: 0

[#permalink]

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.

Please advice if this reasoning is wrong.

Joined: 22 Aug 2005

Posts: 1133

Location: CA

Followers: 1

Kudos [?]: 9 [0], given: 0

[#permalink]

03 Oct 2005, 11:26

jrabenho wrote:

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

[#permalink] 03 Oct 2005, 11:26

		Similar topics	Author	Replies	Last post
Similar Topics:		There are six balls in a black sack. Among those six are	stolyar	7	27 Jun 2003, 01:35
		There are six balls in a black sack. Among those six are	stolyar	10	27 Jun 2003, 02:56
		A person wrote six letters with six envelopes to different	Praetorian	9	17 Sep 2003, 08:29
		Six People are on an elevator that stops at exactly six	mbassmbass04	2	30 Dec 2004, 23:44
	4	six or not six	pradhan	15	14 Dec 2009, 21:49

Six doctors, six nurses and six anesthesiologists are at a

Main navigation

GMAT Resources

Partners

MBA Resources

GMAT ® is a registered trademark of the Graduate Management Admission Council ® (GMAC ®). GMAT Club's website has not been reviewed or endorsed by GMAC.

GMAT Courses

Admissions Consulting

Six doctors, six nurses and six anesthesiologists are at a

Six doctors, six nurses and six anesthesiologists are at a

Main navigation

GMAT Resources

Partners

MBA Resources