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

Question

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

abhishekkpv · Accepted Answer

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..

Bunuel · Answer

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.

voodoochild · Answer

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....

EvaJager · Answer

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!

jlgdr · Answer

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

ConnectTheDots · Answer

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

JeffTargetTestPrep · Answer

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

energetics · Answer

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.

Kinshook · Answer

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.

Total number of ways = (0+1+2+3+4)(0+1+2+3)(0+1+2) = 5*4*3 = 60 ways

IMO A

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards