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Combinations problem - Need expert help

by voodoo_child » Tue Jul 10, 2012 7:31 am
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.

1) 60
2) 1260
3) 24
4) 120
5) 9

OA - A

Any thoughts why it's not B = 9!/(4! 3! 2!)?
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by Brent@GMATPrepNow » Tue Jul 10, 2012 8:03 am
voodoo_child 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.

1) 60
2) 1260
3) 24
4) 120
5) 9

OA - A

Any thoughts why it's not B = 9!/(4! 3! 2!)?
Let's take the task of selecting the balls and break it into stages.

Stage 1: Decide how many blue balls we want.
We can have 0, 1, 2, 3, or 4 blue balls.
So, stage 1 can be accomplished in 5 ways.

Stage 2: Decide how many yellow balls we want.
We can have 0, 1, 2, or 3 yellow balls.
So, stage 2 can be accomplished in 4 ways.

Stage 3: Decide how many green balls we want.
We can have 0, 1 or 2 green balls.
So, stage 3 can be accomplished in 3 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages in 5x4x3 ways (60 ways)

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Brent
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by Brent@GMATPrepNow » Tue Jul 10, 2012 8:07 am
For more information on the FCP, you can watch this free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

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by mmeital » Wed Jul 11, 2012 9:16 am
Brent,

The reason we can use FCP in this question is because each set of colored balls is IDENTICAL correct? How would you solve this question if we weren't told this bit of information?
I originally started out (without taking the identicl into account):
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

Do I now add up the options = 29 ways? that seems too low.
Do I multiply the options = 1X4X8X16= 512 ways?
Should I multiply all the combinations:
0 balls - 1 options
Green Balls = 1X2X1 = 2 ways
Yellow balls = 1x3x3x1 = 9 ways
Blue Balls = 1x4x6x4x1 = 96 ways
For total of 1x2x9x96=1728 ways?

Thanks in advance!

Matan
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by Brent@GMATPrepNow » Wed Jul 11, 2012 12:16 pm
mmeital wrote:Brent,
The reason we can use FCP in this question is because each set of colored balls is IDENTICAL correct? How would you solve this question if we weren't told this bit of information?
I originally started out (without taking the identicl into account):
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

Do I now add up the options = 29 ways? that seems too low.
Hmmm, this solution is assuming that the selected balls must be the same color.

mmeital wrote:Brent,
Do I multiply the options = 1X4X8X16= 512 ways?
Yes! (if we say that the balls of a certain colour are NOT identical)

Here's what you're doing.

Aside: In the next post, I'll show you a much faster solution to this side question.

You're taking the task of selecting the balls, and you're breaking it into stages.

Stage 1: select some green balls
As you've shown, this can be accomplished in 4 ways

Stage 2: select some yellow balls
As you've shown, this can be accomplished in 8 ways

Stage 3: select some blue balls
As you've shown, this can be accomplished in 16 ways

So, the total # of ways to select the ball = 4x8x16 = 512 (if we say that the colored balls are NOT IDENTICAL.

I hope that helps.

Cheers,
Brent
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by Brent@GMATPrepNow » Wed Jul 11, 2012 12:22 pm
mmeital wrote:Brent,
The reason we can use FCP in this question is because each set of colored balls is IDENTICAL correct? How would you solve this question if we weren't told this bit of information?
In other words, what is the answer if the colored balls are NOT identical?
Here's the quick approach.

First recognize that, if the colored balls are NOT identical, then we basically have 9 different balls. In how many different ways can we select some (or even none) of these balls?

Let's take the balls and call them ball #1, ball #2, ball #3, . . . ball #9

Take the task of selecting the balls and break it into 9 stages.

Stage 1: determine whether or not ball #1 is to be included in the selection.
This stage can be accomplished in 2 ways (select it or don't select it)

Stage 2: determine whether or not ball #2 is to be included in the selection.
This stage can be accomplished in 2 ways (select it or don't select it)

Stage 3: determine whether or not ball #3 is to be included in the selection.
This stage can be accomplished in 2 ways.

.
.
.
Stage 9: determine whether or not ball #9 is to be included in the selection.
This stage can be accomplished in 2 ways.

By the FCP, the total number of ways to select some (or none) of the balls (if the colored balls are NOT identical) is equal to 2x2x2x2x2x2x2x2x2 = 2^9 = 512

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by mmeital » Thu Jul 12, 2012 6:57 am
Thanks!

This really helps a lot. If the balls are not identical, does that mean that colors don't matter, or can we assume that same colored balls are identical?

Can you explain please why !9 is not the answer for selecting 9 balls, if balls are not identical?

Thanks again,

Matan
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by Brent@GMATPrepNow » Thu Jul 12, 2012 7:04 am
mmeital wrote:Thanks!

This really helps a lot. If the balls are not identical, does that mean that colors don't matter, or can we assume that same colored balls are identical?

Can you explain please why !9 is not the answer for selecting 9 balls, if balls are not identical?

Thanks again,

Matan
We'd use 9! to determine the number of ways to select 9 different balls if the order of the selection mattered. Similarly, we'd use 9! to determine the number of ways to arrange 9 different balls in a line.

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by voodoo_child » Sat Jul 21, 2012 7:34 am
Brent,
I believe that the answer to the last post would be greater than 9!. Why?

LEt's say that there are only three different balls - Red, Green and Blue.

Possible ordered arrangements =

No No No = 1 arrangement
_ _ A (3 times for different color X3 times for different position) =9 [For instance, - - A; - A - ; A - -; repeated for three balls. = 9]

_ B A (6 times for choosing 2 balls out of 3 balls X 3 times for different positions)=18
A B C (This = 6)

Total = 34 > 3! Correct?

Why is it so? X! assumes that one would choose ALL balls in a different arrangemnet. IT doesn't include combinations when one doesn't choose ALL the balls. Correct? However, I don't know how I will generalize this problem to 9 different balls. Can you please help me?

Please let me know your thoughts.

Thanks
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by Brent@GMATPrepNow » Sat Jul 21, 2012 7:47 am
voodoo_child wrote:Brent,
I believe that the answer to the last post would be greater than 9!. Why?
I think we now have a problem where there are several different questions and answers on this thread. My 9! comment was a response to the question " Why is !9 not the answer?" I was just suggesting when a person might use 9! (that is the number of ways to arrange 9 different objects in a row).

voodoo_child wrote:X! assumes that one would choose ALL balls in a different arrangemnet. IT doesn't include combinations when one doesn't choose ALL the balls. Correct? However, I don't know how I will generalize this problem to 9 different balls. Can you please help me?
If we treat all 9 balls as different, and we can select any number of balls, and the order of the selection doesn't matter, then the answer is 2^9.

I'll repost my solution here:

First recognize that, if the colored balls are NOT identical, then we basically have 9 different balls. In how many different ways can we select some (or even none) of these balls?

Let's take the balls and call them ball #1, ball #2, ball #3, . . . ball #9

Take the task of selecting the balls and break it into 9 stages.

Stage 1: determine whether or not ball #1 is to be included in the selection.
This stage can be accomplished in 2 ways (select it or don't select it)

Stage 2: determine whether or not ball #2 is to be included in the selection.
This stage can be accomplished in 2 ways (select it or don't select it)

Stage 3: determine whether or not ball #3 is to be included in the selection.
This stage can be accomplished in 2 ways.

.
.
.
Stage 9: determine whether or not ball #9 is to be included in the selection.
This stage can be accomplished in 2 ways.

By the FCP, the total number of ways to select some (or none) of the balls (if the colored balls are NOT identical) is equal to 2x2x2x2x2x2x2x2x2 = 2^9 = 512

Cheers,
Brent
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by voodoo_child » Sat Jul 21, 2012 8:55 am
Brent,
I agree with every single word in your post. I thought that the original question was a bit different. For instance, say, there are 9 balls of different colors. A kid can choose as many number of balls as he wants so that he could arrange them in a single line. He could also decide to choose no balls. What would be the solution in that case ?

I could find the solution, as described above, for 3 balls. I literally crashed while calculating the answer for 9 balls. Can you please help me? Is there any shortcut method?

Thanks
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by Brent@GMATPrepNow » Sat Jul 21, 2012 9:12 am
voodoo_child wrote:Brent,
I agree with every single word in your post. I thought that the original question was a bit different. For instance, say, there are 9 balls of different colors. A kid can choose as many number of balls as he wants so that he could arrange them in a single line. He could also decide to choose no balls. What would be the solution in that case ?

I could find the solution, as described above, for 3 balls. I literally crashed while calculating the answer for 9 balls. Can you please help me? Is there any shortcut method?

Thanks
Hmm, so we're saying that order matters AND we have 9 distinct balls AND we can select from 0 to all 9 balls. That's pretty complex, in fact it's probably out of scope for the GMAT (unless I'm missing something basic).

I think you need to treat each case separately and then add the results from all of the cases.

case 1: Select 0 balls. This can be accomplished in 1 way.
case 2: Select 1 ball. This can be accomplished in 9 ways.
case 3: Select 2 balls. This can be accomplished in 9x8 ways (9 options for the first ball and 8 options for the 2nd ball).
case 4: Select 3 balls. This can be accomplished in 9x8x7 ways
case 5: Select 4 balls. This can be accomplished in 9x8x7x6 ways
.
.
.
case 10: Select all 9 balls. This can be accomplished in 9x8x7...x2x1 ways (9! ways)

Add up all 10 cases. . . . WAYYY too much work. Of course, there might be a shortcut, but I don't see it just yet.

Cheers,
Brent
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by voodoo_child » Sun Jul 29, 2012 11:27 am
Brent@GMATPrepNow wrote:
mmeital wrote:Brent,
The reason we can use FCP in this question is because each set of colored balls is IDENTICAL correct? How would you solve this question if we weren't told this bit of information?
I originally started out (without taking the identicl into account):
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

Do I now add up the options = 29 ways? that seems too low.
Hmmm, this solution is assuming that the selected balls must be the same color.
Brent - I believe that you meant "different color"..This is a typo. Correct? (I have underlined with blue)....
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by Brent@GMATPrepNow » Sun Jul 29, 2012 2:32 pm
voodoo_child wrote:
Brent@GMATPrepNow wrote:
mmeital wrote:Brent,
The reason we can use FCP in this question is because each set of colored balls is IDENTICAL correct? How would you solve this question if we weren't told this bit of information?
I originally started out (without taking the identicl into account):
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

Do I now add up the options = 29 ways? that seems too low.
Hmmm, this solution is assuming that the selected balls must be the same color.
Brent - I believe that you meant "different color"..This is a typo. Correct? (I have underlined with blue)....
No, I meant "same color." Mmeital's solution (above) was operating on the idea that the selected balls had to be the same color. So, if someone chose 3 balls, all 3 balls had to the the same color (in Mmeital's solution). In other words, you couldn't select 1 yellow ball, 1 blue ball and 1 green ball. That is not to say that balls of the same color are identical. Mmeital's top solution assumes that, within balls of the same color, each ball is different.

Cheers,
Brent
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by voodoo_child » Sun Jul 29, 2012 2:42 pm
Yes, Brent you are absolutely correct. I meant different ball of the same color (may be the balls have unique scratches) and not the ball of different color. :( that was a typo on my side :(
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