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A bag has 4 blue, 3 yellow and 2 green balls. The balls of [#permalink]
30 Sep 2012, 08:13
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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
Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of [#permalink]
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.. _________________
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! _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of [#permalink]
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.
Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of [#permalink]
19 Jul 2015, 19:05
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