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18 children, including 9 girls and 9 boys, are to be choses

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18 children, including 9 girls and 9 boys, are to be choses [#permalink] New post 18 Sep 2009, 13:09
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18 children, including 9 girls and 9 boys, are to be choses to form a team of 4 people. Players are selected one at a time. How many teams are possible with exactly one girl on the team?
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Re: Probability question [#permalink] New post 18 Sep 2009, 14:07
you have four places inthe team to fill:

- - - -

you have to fil exactly one with a girl and rest 3 with boys:

Suppose you want G B B B

for the girl's place (1st team member's place) you have 9 choices.

for 2nd place you have 9 choice of boy
for 3rd place you have 8 choices of boy (since one boy is already used for 2nd place)
and similarly 4th

SO you have total of: G B B B
9 x 9 x 8 x 7 = _ so many choices.

but wait you need to take into consideraitons repeats also you can have GBBB or BBBG or BBGB ... and so on.

you you have to divide my (# of places)! i,e 4!

therefore ans: 9x9x8x7/4!
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Re: Probability question [#permalink] New post 18 Sep 2009, 16:29
9 x 9 x 8 x 7 = _ so many choices.

but wait you need to take into consideraitons repeats also you can have GBBB or BBBG or BBGB ... and so on.

you you have to divide my (# of places)! i,e 4!

therefore ans: 9x9x8x7/4!

yeeeeeeeaaahhhh you made sense until you divided by 4.... but 4,536 is the right answer as far as I can tell
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Re: Probability question [#permalink] New post 18 Sep 2009, 18:59
vannbj wrote:
9 x 9 x 8 x 7 = _ so many choices.

but wait you need to take into consideraitons repeats also you can have GBBB or BBBG or BBGB ... and so on.

you you have to divide my (# of places)! i,e 4!

therefore ans: 9x9x8x7/4!

yeeeeeeeaaahhhh you made sense until you divided by 4.... but 4,536 is the right answer as far as I can tell


You need to divide by 4! because the order in which we place the boys and girls doesn't matter those 4 team player will make that 1 unique team no matter if you count them as B1 G1 B2 B3 or B1 B2 G1 B3 or any other way.
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Re: Probability question [#permalink] New post 18 Sep 2009, 20:48
OR
9*9C3 which gives the total number of 'unique' boy and girl combinations ( say names are unique ) G1,B1,B2,B3
9*9C3/4! gives the combinations ( out of those unique combinations ) in which only gender is considered G1,B1,B2,B3 becomes GBBB. ( as we do not need to differentiate between each boy (B1=B2) or each girl (G1=G2) )
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Re: Probability question [#permalink] New post 18 Sep 2009, 20:59
I did it this way:

9C3*9 = 756.

Is this approach correct?
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Re: Probability question [#permalink] New post 18 Sep 2009, 22:04
hgp2k wrote:
I did it this way:

9C3*9 = 756.

Is this approach correct?

This is what I mentioned in my earlier post :) If we use this approach we end up having unique name combinations...but the question asks for 'any' boy or 'any' girl, we do not have to differentiate between b1 and b2 and b3...eg. (G1 B1 B2 B3) is the same as (G1 B2 B1 B3) or (G1 B3 B2 B1)..so in order to ignore such repeating combinations we have to divide by 4!
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Re: Probability question [#permalink] New post 25 Sep 2009, 13:17
Economist wrote:
hgp2k wrote:
I did it this way:

9C3*9 = 756.

Is this approach correct?

This is what I mentioned in my earlier post :) If we use this approach we end up having unique name combinations...but the question asks for 'any' boy or 'any' girl, we do not have to differentiate between b1 and b2 and b3...eg. (G1 B1 B2 B3) is the same as (G1 B2 B1 B3) or (G1 B3 B2 B1)..so in order to ignore such repeating combinations we have to divide by 4!



9C3*9 / !4 = 31.5. How a number of possible arrangements be a non-integer number?
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Re: Probability question [#permalink] New post 07 Apr 2010, 07:36
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This is what I mentioned in my earlier post If we use this approach we end up having unique name combinations...but the question asks for 'any' boy or 'any' girl, we do not have to differentiate between b1 and b2 and b3...eg. (G1 B1 B2 B3) is the same as (G1 B2 B1 B3) or (G1 B3 B2 B1)..so in order to ignore such repeating combinations we have to divide by 4!



I think you're only supposed to divide by 3! to remove the boys repeating combinations.

There are two approaches:

Combinatronics:

9C1 - Choose a girl from the 9 available
9C3 - Choose three boys from the 9 available

9C1*9C3 = 756 arrangements [repetitions already discounted by the combinations formula]


Counting:

9 options of girls for first place
9 options of boys for second place
8 options of boys for third place
7 options of boys for fourth place
3! discount ordering possibilities in the boys places

(9*9*8*7) / 3!


[N.B. This is the same as 9C1*9C1*8C1*7C1 / 3!]
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Re: Probability question [#permalink] New post 07 Apr 2010, 11:32
Can you post the answer OP (or get Bunuel in here, which is as good as the OA :p)?

I got 3024

9C1*4C1*9C3

9C1: number of ways to select 1 girl from 9
4C1: number of ways the girl can be arranged in between the boys (thats what she said)
9C3: number of ways to choose the remaining 3 boys
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Re: Probability question [#permalink] New post 08 Apr 2010, 02:30
Expert's post
rlevochkin wrote:
18 children, including 9 girls and 9 boys, are to be choses to form a team of 4 people. Players are selected one at a time. How many teams are possible with exactly one girl on the team?


This is badly constructed question. If it were: there are 9 girls and 9 boys. How many different teams of 4 can be formed from them, with exactly one girl?

9C1*9C3=756.
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Re: Probability question [#permalink] New post 08 Apr 2010, 02:32
Bunuel wrote:
rlevochkin wrote:
18 children, including 9 girls and 9 boys, are to be choses to form a team of 4 people. Players are selected one at a time. How many teams are possible with exactly one girl on the team?


This is badly constructed question. If it were: there are 9 girls and 9 boys. How many different teams of 4 can be formed from them, with exactly one girl?

9C1*9C3=756.


Why shouldn't 756 be multiplied by 4C1? The girl can be chosen first, second third or fourth....
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Re: Probability question [#permalink] New post 08 Apr 2010, 02:41
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Expert's post
nickk wrote:
Bunuel wrote:
rlevochkin wrote:
18 children, including 9 girls and 9 boys, are to be choses to form a team of 4 people. Players are selected one at a time. How many teams are possible with exactly one girl on the team?


This is badly constructed question. If it were: there are 9 girls and 9 boys. How many different teams of 4 can be formed from them, with exactly one girl?

9C1*9C3=756.


Why shouldn't 756 be multiplied by 4C1? The girl can be chosen first, second third or fourth....


It's a team. Does the relative position of members in the team matter?

Consider the easier question: 2 girls G1, G2 and 3 boys B1, B2 and B2. How many different teams of 3 can be formed with exactly one girls?

2C1*3C2=6.
G1, B1, B2, here B1, G1, B2 is the same team;
G1, B1, B3;
G1, B2, B3;
G2, B1, B2;
G2, B1, B3;
G2, B2, B3.
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Re: Probability question [#permalink] New post 09 Apr 2010, 00:56
Players are selected one at a time. - This sentence has no bearing on the outcome/answer.

For those of you who are bent upon dividing by 4! you should be using P(n,r) and not C(n,r). The nCr formula in itself takes cares of clubbing (B1, B2, B3 and their other combinations) into B,B,B.

The answer is:
No. of ways of selecting the only 1 girl = 9C1 = 9
No. of ways of selecting (NOT arranging) the 3 boys = 9C3
If the question differentiates between the boys, then this part becomes 9P3.

Coming back, you need to choose 1 girl AND 3 boys, so (9 x 9C3) = 756
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Re: Probability question [#permalink] New post 09 Apr 2010, 12:33
lprassanth and Bunuel are right.
I got 9C1*9C3=756 as well. It is a team; any order of selection will do.
Re: Probability question   [#permalink] 09 Apr 2010, 12:33
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