In how many ways can we put 4 different balls in 3 different boxes whe

Question

In how many ways can we put 4 different balls in 3 different boxes when any box can contain any number of balls?

A. 80
B. 81
C. 64
D. 63
E. 82

Answer: B.

Can someone please tell me why 4x4x4=64 is not the right answer?
There are 4 ways to fill the 1st box, and 4 ways to the 2nd box and 4 ways to fill the 3rd box..since it says that any box can contain any number of balls..i assumed that repetition is allowed.

Kindly explain.

Thanks alot!!!

BrentGMATPrepNow · Accepted Answer

Take the task of distributing the 4 different balls and break it into  stages .

Stage 1 : Select a box for the 1st ball to go into.
There are 3 available boxes, so we can complete stage 1 in  3  ways

Stage 2 : Select a box for the 2nd ball to go into.
There are 3 available boxes, so we can complete stage 2 in  3  ways

Stage 3 : Select a box for the 3rd ball to go into.
There are 3 available boxes, so we can complete stage 3 in  3  ways

Stage 4 : Select a box for the 4th ball to go into.
There are 3 available boxes, so we can complete stage 4 in  3  ways

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus distribute all 4 balls) in  (3)(3)(3)(3)(4)  ways (= 81 ways)

Answer:  B

RELATED VIDEO
 https://www.youtube.com/watch?v=w5d5bvCRyd4

BrentGMATPrepNow · Answer

"  There are 4 ways to fill the 1st box  "
This assumes that the box needs to be filled. 
No requirement for any one box to contain ANY balls.

On the other hand, every ball needs to be placed somewhere. 
So, we need to examine the number of ways to place each ball. See my solution above.

Cheers,
Brent

chetan2u · Answer

Just adding my response from Chat on why 3^4 and not 4^3. May help few.

Now, about 4^3 ways. This comes from any of the FOUR balls in boxes. Say balls are A, B , C and D, while boxes are X, Y and Z. For A in X, any of A, B, C, and D in Y and again any of A, B, C, and D in Z. But can the same ball A be there in all three boxes.....NO.

But in case of 3^4, can same box X, have all four balls...YES. I hope it clears the query on why 3^4 and not 4^3.

Solution:

(I) Calculate each way.. 
  1,0,3:   first 3! ways in which 1, 0 and 3 balls can be placed. Next these 3 balls will have to be selected in 4C3 ways. So Box 1-1, Box 2-0 and Box3 -3 will not be one way but 4C3 ways as the 3 in box 3 can be selected in 4C3 ways. Thus, total ways for (1,0,3) will be 3!*4C3 or 24.
  1,1,2:   Choosing the box with 2 balls can be in 3C1 or 3 ways. Next selecting/choosing 2 balls from 4 will be in 4C2 or 6 ways and then selecting one ball from the remaining two balls in 2C1 way and the remaining ball will go in the final box in 1C1, so .....3*4C2*2C1*1C1 = 3*6*2*1 = 36 ways 
  0,0,4:   No unique values as all are in same box, but box can change in 3C1 ways, so 3 ways.
  0,2,2:   Choose boxes in 3!/2! or 3 ways. Then choose 2 balls in 4C2 ways and the remaining 2 will automatically come out in 1 way, so 3*4C2*2C2 = 3*6*1 or 18 ways.
 Total: 24+36+3+18 = 81 ways or 3^4 ways

(II)  There are 4 balls and each can go in any of the three boxes, so 3*3*3*3 or 3^4 ways , that is 81 ways

msk0657 · Answer

NOTE : The number of permutations/arrangments of n things , taken r at ta time when each item may be repeated once, twice...up to r times in any arrangement is  n^r  ways.

The first box can be filled in n ways, the second box can be filled in n ways(even though the first box is filled with one item, the same item can be used for filling the second box also because repetition is allowed), the third box can also be filled in n ways...

The rth box can be filled in n ways..

Now all the r boxes together can be filled in (n*n*n*....r times ) ie.  n^r  ways.

So  3^4  ways..

Hope it clears..

onamarif · Answer

Since there are  3  boxes, we have

3  possibilities for the first ball
 3  possibilities for the second ball
 3  possibilities for the third ball
 3  possibilities for the fourth ball

So, total number of possibilities is  3*3*3*3 = 81

stne · Answer

How could people not give Kudos for such valuable information! Members please appreciate valuable contribution such as this one by giving Kudos! Thanks  GMATPrepNow .

vaibhav1221 · Answer

Another way to think about it by considering the number of balls in each box.

So situation 1 has 3 different possibilities. Similarly, situation 2, 3, and 4 have  3 different possibilities each. 
 
 Situation 2
Box 1: 3 balls
Box 2: 1 balls
Box 3: 0 balls

Situation 3
Box 1: 2 balls
Box 2: 2 balls
Box 3: 0 balls

Situation 4
Box 1: 2 balls
Box 2: 1 balls
Box 3: 1 balls

Thus the total number of ways in which 4 different balls can be put in 3 different boxes when any box can contain any number of balls is 
  3 x 3 x 3 x 3 = 81

exc4libur · Answer

given: 4 dif balls, 3 dif boxes, any number per box;

4•4•4=81

{400}={3!/2!}•4c4=3  ("/2!" because 00 are identical)
 {310}={3!}•4c3=6•4=24 
 {220}={3!}•4c2/2!=6•3=18  ("/2!" because 22 is double counting)
 {211}={3!}•4c2=6•6=36 
 {total}=3+24+18+36=81

Answer (B)

VIVA1060 · Answer

Hi  Bunuel   chetan2u ,

I was trying to use the divider method that  dabaobao  has used in  https://gmatclub.com/forum/in-how-many-ways-can-5-apples-identical-be-distributed-among-4-child-206279.html

Like this: BB|B|B. Therefore, 6!/(2!*4!). However, I am not getting the answer. Can you please advice where I am going wrong?

chetan2u · Answer

That formula is for identical things but here we have different things.

Say we have 3 boxes A,B and C and ball w,x,y, and z
ThE formula gives A- 3 boxes, B- 1 box and C-0 box as one way. 
But the actuals where balls are different 
A- wxy, B-z, C-0 and A-xyz, B-w,C-0 will be treated differently.

varun325 · Answer

Hey! can you help me in understanding this case :   {220}={3!}•4c2/2!=6•3=18  ("/2!" because 22 is double counting) 
 why you divided by 2! in the case 220. The cases will be 220 , 202, 022 and all these are different.
Thanks in advance

bumpbot · Answer

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In how many ways can we put 4 different balls in 3 different boxes whe

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