Problem Solving

nahid078 wrote:In how many ways 5 different balls can be distributed in three boxes?

This question is a little ambiguous.
I'm assuming that the 3 boxes are also different (call them boxes A, B and C)
Let's also call the balls V, W, X, Y and Z

So, take the task distributing balls and break it into stages.

Stage 1: place ball V in a box
There are 3 boxes, so, we can complete stage 1 in 3 ways

Stage 2: place ball W in a box
There are 3 boxes, so, we can complete stage 2 in 3 ways

Stage 3: place ball X in a box
There are 3 boxes, so, we can complete stage 3 in 3 ways

Stage 4: place ball Y in a box
There are 3 boxes, so, we can complete stage 4 in 3 ways

Stage 5: place ball Z in a box
There are 3 boxes, so, we can complete stage 5 in 3 ways

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus distribute all 5 balls) in (3)(3)(3)(3)(3) ways ([spoiler]= 243 ways[/spoiler])

--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent

nahid078 wrote:But if it is said that, one box can contain only one ball. Then How many ways can we put three balls in three boxes?

5*4*3= 60

Would it be the answer?

Thanks

Sorry, I think you need to clarify that restriction. Are you saying that one box must have exactly one ball, and the other 2 boxes can have any number of balls?

If so, the answer is (3)(2)(2)(2)(2)

If you mean something else, please tell me.

Cheers,
Brent

nahid078 wrote:No i said every box can contain only one ball. Thus last two balls will not be in the box.

Then yes, the answer is (5)(4)(3)

Choose a ball to go in box A: 5 balls to choose from
Choose a ball to go in box B: 4 balls remaining to choose from
Choose a ball to go in box C: 3 balls remaining to choose from

Total = (5)(4)(3)

Cheers,
Brent

Amrabdelnaby wrote:Hi Brent,

Thank you for the answer; however I have a little question.

Why did you distribute the boxes on the balls and not vice versa?

I thought about distributing the balls on the boxes

so i thought that we have three boxes, for the first one we can put one of 5 balls and for the second we can put one of four balls and for the third we can put one of three.

leaving me with the following calculation 5x4x3 = 60 ways.

Why is this wrong?

Brent@GMATPrepNow wrote:

nahid078 wrote:In how many ways 5 different balls can be distributed in three boxes?

This question is a little ambiguous.
I'm assuming that the 3 boxes are also different (call them boxes A, B and C)
Let's also call the balls V, W, X, Y and Z

So, take the task distributing balls and break it into stages.

Stage 1: place ball V in a box
There are 3 boxes, so, we can complete stage 1 in 3 ways

Stage 2: place ball W in a box
There are 3 boxes, so, we can complete stage 2 in 3 ways

Stage 3: place ball X in a box
There are 3 boxes, so, we can complete stage 3 in 3 ways

Stage 4: place ball Y in a box
There are 3 boxes, so, we can complete stage 4 in 3 ways

Stage 5: place ball Z in a box
There are 3 boxes, so, we can complete stage 5 in 3 ways

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus distribute all 5 balls) in (3)(3)(3)(3)(3) ways ([spoiler]= 243 ways[/spoiler])

--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent

Which question are you referring to?
In my solution here, I am distributing the balls into the boxes.

Cheers,
Brent