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A bag has 4 blue, 3 yellow and 2 green balls. The balls of

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A bag has 4 blue, 3 yellow and 2 green balls. The balls of [#permalink] New post 30 Sep 2012, 08:13
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44% (02:40) correct 56% (01:52) wrong based on 71 sessions
A bag has 4 blue, 3 yellow and 2 green balls. The balls of the same color are identical. In how many ways can a child picks ball(s) out of the bag? He could even decide to pick ZERO balls.

A. 60
B. 1260
C. 24
D. 120
E. 9

Can someone please explain why the answer isn't :

Option 1: 0 balls - 1 way
2. Green balls: (2C0=1)+(2C1=2)+(2C2=1)= 4 ways
3. Yellow Balls: (3C0=1)+(3C1=3)+(3C2=3)+(3C3=1)=8 ways
4. Blue Balls: (4C0=1)+(4C1=4)+(4C2=6)+(4C3=4)+(4C4=1)= 16 ways

Total => multply all the numbers abvoe => 512

I am a bit confused.
[Reveal] Spoiler: OA
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Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of [#permalink] New post 30 Sep 2012, 10:24
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With the blue ball there are 5 scenarios-->
0 blue ball being selected
1 blue ball being selected
2 blue ball being selected
3 blue ball being selected
4 blue ball being selected

And for each of these cases there is only one possibilty as they are identical.

So there are 5 ways to select the blue ball
Same is the case with 3 yellow (for which we have 4 possibilties) and 2 green balls(we have 3 possibilties)


so in total 5*4*3 = 60 possibilties..
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Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of [#permalink] New post 30 Sep 2012, 11:04
Ok. But why can't we do Green balls: (2C0=1)+(2C1=2)+(2C2=1)= 4 ways ? See this thread : a-jar-contains-6-magenta-balls-139813.html

The balls are also identical in the above question. I am confused between these two threads....
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Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of [#permalink] New post 30 Sep 2012, 12:24
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voodoochild wrote:
Ok. But why can't we do Green balls: (2C0=1)+(2C1=2)+(2C2=1)= 4 ways ? See this thread : a-jar-contains-6-magenta-balls-139813.html

The balls are also identical in the above question. I am confused between these two threads....


All the Green balls are identical. We don't care which one(s) was(were) chosen, only how many. We are interested in the final "statistics": how many Green balls were chosen? We have 3 possibilities: 0, 1, or 2.
2C1 = 2, it's true, but choosing either Green ball, will have the same statistical result: one Green ball was chosen, and we don't care about which one of the two Greens, because they are absolutely identical!
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Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of [#permalink] New post 16 Apr 2014, 13:23
I tried this problem using anagram grid

9! / 4! 3 ! 2! = 1260

Does anyone know what i'm missing to get to 60 as an answer here?

Thanks!
Cheers
J :)
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Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of [#permalink] New post 18 Apr 2014, 10:10
Expert's post
jlgdr wrote:
A bag has 4 blue, 3 yellow and 2 green balls. The balls of the same color are identical. In how many ways can a child picks ball(s) out of the bag? He could even decide to pick ZERO balls.

A. 60
B. 1260
C. 24
D. 120
E. 9

I tried this problem using anagram grid

9! / 4! 3 ! 2! = 1260

Does anyone know what i'm missing to get to 60 as an answer here?

Thanks!
Cheers
J :)


We are not asked to find the number of arrangements of 4 blue, 3 yellow and 2 green balls, which would be 9!/(4!3!2!).

We are asked to find the number of selections from 4 blue, 3 yellow and 2 green balls, when there can be any number of balls selected from 0 to 9.

A bag has 4 blue, 3 yellow and 2 green balls. The balls of the same color are identical. In how many ways can a child picks ball(s) out of the bag? He could even decide to pick ZERO balls.

A. 60
B. 1260
C. 24
D. 120
E. 9

In a selection there can be:
0, 1, 2, 3, or 4 blue balls, so 5 cases;
0, 1, 2, or 3 yellow balls, so 4 cases;
0, 1, or 2 green balls, so 3 cases.

Total 5*4*3=60 selections.

Answer: A.

Hope it's clear.
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Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of [#permalink] New post 18 Apr 2014, 11:34
please go thru below link:

math-combinatorics-87345.html#p785179

You can solve in seconds: (4+1)*(3+1)*(2+1) = 5*4*3 = 60
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Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of   [#permalink] 18 Apr 2014, 11:34
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