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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 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
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Originally posted by voodoochild on 30 Sep 2012, 09:13.
Last edited by pushpitkc on 25 Feb 2018, 06:19, edited 1 time in total.
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Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of
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30 Sep 2012, 11:24
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
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30 Sep 2012, 12:04
Ok. But why can't we do Green balls: (2C0=1)+(2C1=2)+(2C2=1)= 4 ways ? See this thread : ajarcontains6magentaballs139813.htmlThe 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
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30 Sep 2012, 13:24
voodoochild wrote: Ok. But why can't we do Green balls: (2C0=1)+(2C1=2)+(2C2=1)= 4 ways ? See this thread : ajarcontains6magentaballs139813.htmlThe 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
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16 Apr 2014, 14:23
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
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18 Apr 2014, 11:10
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
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18 Apr 2014, 12:34
please go thru below link: mathcombinatorics87345.html#p785179You 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
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18 Aug 2018, 19:32
voodoochild 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 For the blue balls, he can choose 0, 1, 2, 3 or all 4 blue balls. So he has 5 choices for selecting the number of blue balls. Similarly, for the yellow balls, he can choose 0, 1, 2 or all 3 yellow balls. So he has 4 choices for selecting the number of yellow balls. Last but not least, for the green balls, he can choose 0, 1, or all 2 green balls. So he has 3 choices for selecting the number of green balls. Therefore, the total number of ways he can choose the balls is: 5 x 4 x 3 = 60 Answer: A
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Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of
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05 Jul 2019, 09:32
voodoochild 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 JeffTargetTestPrep does this explanation make sense? The distinction here is between the number of selections of "identical things into identical groups" and the number of ways to select "identical things into identical groups" (i.e. that all the balls of the same color are the same and the groups they are put in are the same, so we are not at all concerned with whether we select Blue 1, Blue 2, Blue 3, or Blue 4 nor are we concerned with the group we put them in comparison with others (if we put B1 and G1 together, or G2 and B2 it all counts as the same thing). If either of these things were considered the number would be much larger. We instead say that the ways to select Blue are from 0 to 4, inclusive ... (40)+1 = 5 ways to select a blue ball. Meaning that we can choose 0 to 4 blue balls, regardless of what else is going on (because again, we don't care about the relation to other blue balls or other balls in the group). We multiply 5*4*3 = 60 because that's the number of WAYS we can put balls together, below are all the possibilities: 9 WAYS TO CHOOSE 1 TYPEB, BB, BBB, BBBB, G, GG Y, YY, YYY 26 WAYS TO CHOOSE 2 TYPESBG, BBG, BBBG, BBBBG, BGG, BBGG, BBBGG, BBBBGG, BY, BBY, BBBY, BBBBY, BYY, BBYY, BBBYY, BBBBYY, BYYY, BBYYY, BBBYYY, BBBBYYY, YG, YGG, YYG, YYGG YYYGG, YYYGG 24 WAYS TO CHOOSE 3 TYPESBYG, BBYG, BBBYG, BBBBYG, BYYG, BBYYG, BBBYYG, BBBBYYG, BYYYG, BBYYYG, BBBYYYG, BBBBYYYG, BYGG, BBYGG, BBBYGG, BBBBYGG, BYYGG, BBYYGG, BBBYYGG, BBBBYYGG, BYYYGG, BBYYYGG, BBBYYYGG, BBBBYYYGG, That adds up to 59, what's left is 1 WAY TO CHOOSE 0 TYPES so + 1 = 60.



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Re: A bag has 4 blue, 3 yellow and 2 green balls. The balls of
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08 Jul 2019, 00:56
Bunuel wrote: 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. Bunnel, Please help me out here! Question asked 'how many ways'. it implies {first blue then red} and f{first red then blue} are two different ways. so, 9!/4!3!2! should be correct because it gives no. of arrangements.




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