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

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!

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

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

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

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 ... (4-0)+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 TYPE
B, BB, BBB, BBBB,
G, GG
Y, YY, YYY

26 WAYS TO CHOOSE 2 TYPES
BG, 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 TYPES
BYG, 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.

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.

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