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Q5: From a group of 3 boys and 3 girls, 4 children are to

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Manager
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Q5: From a group of 3 boys and 3 girls, 4 children are to [#permalink] New post 08 Jun 2007, 08:11
00:00
A
B
C
D
E

Difficulty:

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Question Stats:

100% (01:01) correct 0% (00:00) wrong based on 0 sessions
Q5:
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3

Q6:
A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?

A. 5
B. 6
C. 7
D. 8
E. 9
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 [#permalink] New post 08 Jun 2007, 20:22
5)
6c4=15 total ways to select boys and girls
3c2 = 3 ways to select 2 boys and 3c2 = 3 ways to select 2 girls
3*3 = 9 ways to select 3 boys and 3 girls
9/15 = 3/5

6)
I wrote each option out:
office a - all 3 emplooyees, 2 employees or 1 employee. there are 3 options here. office b can also have each of the same arrangments. 3*2 = 6
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 [#permalink] New post 09 Jun 2007, 16:29
ggarr wrote:
5)
6c4=15 total ways to select boys and girls
3c2 = 3 ways to select 2 boys and 3c2 = 3 ways to select 2 girls
3*3 = 9 ways to select 3 boys and 3 girls
9/15 = 3/5

6)
I wrote each option out:
office a - all 3 emplooyees, 2 employees or 1 employee. there are 3 options here. office b can also have each of the same arrangments. 3*2 = 6



Office can also be empty :) hence answer = 8 !

Also, can someone point out the method to solve this problem in a more systematic way than counting ?

say the q was .. to place 1000 employees in 20 offices !
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 [#permalink] New post 09 Jun 2007, 18:36
Agree with 3/5 on Q1

I think that answer to question 2 is 7 (Option C).

Way to arrange 3 employees in 2 offices = 3C2 = 3
Way to arrange 3 employees in 1 office = 3C1 = 3
Way in which both offices are empty = 1 so 3+3+1 = 7 ?
  [#permalink] 09 Jun 2007, 18:36
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