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# If a committee of 3 people is to be selected from among 5

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If a committee of 3 people is to be selected from among 5 [#permalink]

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09 Aug 2006, 23:28
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

3. If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

20
40
50
80
120

19. In a survey of 200 college graduates, 30 percent said they had received student loans during their college careers, and 40 percent said they had received scholarships. What percent of those surveyed said that they had received neither student loans nor scholarships during their college careers?

(1) 20 percent of those surveyed said that they had received scholarships but no loans.

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09 Aug 2006, 23:53
D 80

Total number of possibilities = 10C3 = 120
Ways to choose 1 couple = 1*1*8 (1 way to select husband, 1 way to select wife and 1 out of the other 8 people) = 8

Hence for 5 couples total # of ways = 8*5

Hence answer = 120 - 40 = 80

D Each one is sufficient

Given SL = 30% = 60 and SC = 40% = 80

ST 1 20% SC but not SL (which is SC only) = 40 people.

SC only + SL only + SC and SL = 40 + 20 + 40 = 100

Hence answer = 200 - 100 = 100

ST 2 gives us 50% SL = 30 received both SC and SL

Hence the other numbers are

SC only = 80 - 30 = 50
SL only = 60 - 30 = 30

Hence answer is 200 - 50 - 30 30 = 90

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09 Aug 2006, 23:56
Q3:

Total cases = 10C3 = 120
Cases when one couple is selected and third person is another one = 5*8 = 40

Q19:
D

L= 60
S = 80
All = L+S-LS+NO(LS)
200 = 60+80-LS + NO(LS)
NO(LS) = LS + 60

St1: ONLY S = 40. So LS = 80-40 = 40: SUFF

St2: LS = L/2 = 30. :SUFF

But answers by each statement are different.
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10 Aug 2006, 00:06
Can u please explain the formula of 10C3 . and also when I should pick Permutation n when combination? I am lost
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10 Aug 2006, 03:09
razrulz wrote:
3. If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

20
40
50
80
120

D

Number of ways 3 people can be chosen from 10 people = 10C3 = 120

Number of ways a married couple can be chosen = 5 ways
Number of ways the third commitee can be chosen = 8C1 = 8ways

Hence number of commitees formed without married couple = 120 - 5*8 = 80

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10 Aug 2006, 03:12
razrulz wrote:
Can u please explain the formula of 10C3 . and also when I should pick Permutation n when combination? I am lost

10C3 = 10!/[(10-3)! * 3!]

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10 Aug 2006, 07:14
razrulz wrote:
Can u please explain the formula of 10C3 . and also when I should pick Permutation n when combination? I am lost

If we have to select "r" things from "n" things then total ways are

nCr = n!/[r! * (n-r)!)

If we have to select AND ARRANGE "r" things from "n" things then total ways are

nPr = n!/(n-r)!

nPr = nCr * r!

Hope this helps.
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10 Aug 2006, 07:29
ps_dahiya wrote:
razrulz wrote:
Can u please explain the formula of 10C3 . and also when I should pick Permutation n when combination? I am lost

If we have to select "r" things from "n" things then total ways are

nCr = n!/[r! * (n-r)!)

If we have to select AND ARRANGE "r" things from "n" things then total ways are

nPr = n!/(n-r)!

nPr = nCr * r!

Hope this helps.

Thnks made a note of it
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10 Aug 2006, 10:29
I solved both problems myself and agree with the solutions provided. However, I am a little concerned that in the second problem (loans vs. scholarships) the answers seem to be different depending on whether you go by ST1 or ST2. This usually does not happen in GMAT...

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10 Aug 2006, 10:46
You are right... atleast MGMAT clearly states that the two statements in DS on the GMAT will not give you inconsistent results.

v1rok wrote:
I solved both problems myself and agree with the solutions provided. However, I am a little concerned that in the second problem (loans vs. scholarships) the answers seem to be different depending on whether you go by ST1 or ST2. This usually does not happen in GMAT...

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12 Aug 2006, 21:26
5 married couple is 10 people
3 can be chosen out of 10 in 10C3 =120 ways

Now, AB CD EF GH IJ
AB is first couple
8 ways
8*5 = 40 ways
80 ways in which a married couple is not in the committee

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12 Aug 2006, 21:26
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# If a committee of 3 people is to be selected from among 5

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