Marshall has 4 different colored balls and 3 different boxes. In how

select 3 balls first then put them into 3 different boxes
\(4c3* !3\) ( !3 - 3 different balls into 3 different boxes)
as one ball is remaining, and it has 3 options
so total= \(4c3*!3*3\)
72
D:)

for 4 colored balls ; in 3 boxes ; 4c3*3! and for 1 ball it can be out in 3c1 ways in either of 3 boxes
total ways
4c3*3!*3c1 = 72
IMO D

Number of ways = 4C3*3! * 3C1 = 4*6*3 = 72

IMO D

Selecting 2 Balls which will be put in 1 Box- 4C2 & Selecting that 1 Box- 3C1. Other 2 Balls will be put in 2 Boxes in 2 ways
Hence no of ways in which 4 balls will be put in 3 boxes are - 4C2 * 3C1 * 2 = 36

Given are four balls and the three boxes which are all different.

You take the first box and add the first ball. Gives you four possibilities. For the second ball you have three possibilities. Now the two remaining balls have to be put in the remaining two boxes. => This gives you 4*3*1*1 = 12 possibilities. As the boxes are individual each box could hold two balls. So you multiply by three. Giving you the total of 36 possibilities.

Solution:

Since no box must be empty and no ball will be left out, one box will contain two balls and two boxes will contain one ball each.

There are 3 choices for the box to contain two balls. For the remaining 2 boxes, 2 balls can be chosen and ordered from a total of 4 balls in 4P2 = 4!/(4-2)! = (4 x 3 x 2!)/2! = 12 ways. The remaining two balls will go in the first box; thus there is only one way we can do this. Therefore, in total, there are 3 x 12 = 36 ways to accomplish this task.

Answer: C

Both the answers are possible - 36 & 72.
Does this mean the question is wrong or I am missing something ?

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The correct way to do this is- we have to select 2 balls first which we will put together in 1 box.
No. of ways to select 2 balls and put them in 1 of a box is: 4C2 * 3

Now, remaining 2 balls can be arranged in 2! ways (into 2 boxes).

Thus, total ways= 4C2 *3 * 2! = 36

Now, The other way to think is (which is actually incorrect) :

Select that 1 extra ball [2,1,1] which is going into 1 of the boxes in : 4C1 ways
put this in 1 of the boxes in : 3 ways
Total ways to do this = 4*3 = 12 ways

Remaining 3 balls can be arranged in 3! ways.

Hence, total ways = 12 * 3! 72 ways ??

Wait !! We are counting the scenarios twice by this method -> because while selecting the combination for that extra 1 ball in one of the boxes, we select one other ball which will accompany it.

For.e.g: Take boxes as 1,2,3 & balls as A,B,C,D

Now, A1B1, C2,D3 Or B1A1, C2, D3 -- are they different or same ??

Do this on pen & paper if still needs clarification.

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Thanks

Say we have four balls: a b c d
boxes : 1 2 3

Since neither no box nor no balls should be left free,
==> we cannot put more than 2 balls ( max) in a box
==> we cannot put only 1 ball in each box
with this information,
let us assume 2 balls together
==> Total no of possibilities 6 ( ab, ac,ad, bc,bd, cd)
now for each combinations above we can have 3 set of balls ( ab, c,d) to fill in 3 boxes : 3! ways
we have total 6 such cases
Hence total possibilities: 6 x 3! = 36

I also got to C but in a different manner:

1) Number of different 2-pairs of balls out of 4 different balls: 4C2, which gives 6
2) Number of combining 3 sets of balls (independently of whether they are paired or not) in 3 boxes: 3!

3!*6=36
IMO C

4 balls: A, B, C, D

3 distinct boxes: we can have the 1st box = 1st place; 2nd box = 2nd place; and 3rd box = 3rd place

If we were to first choose 3 balls out of 4 ——-then for each distinct group of 3 balls arrange them in 3! ways throughout the 3 places ——and then finally place the remaining ball in 1 of 3 boxes

We would end up with:

(4 c 3) * (3!) * (3) = 4 * 6 * 3 = 72 ways

To show why we would be double counting each case we can run through two possibilities:

Case 1: for the step 1, we end up choosing a group of 3 balls that is: (A—B—C) and 1 left over is (D)

There would be 3! = 6 different Ways to arrange them in the 3 places/boxes

A-B-C
A-C-B
etc.

the one left over ball is D - and for each one of these 6 ways, we would place the ball in 3 different places/boxes.

Let’s just look at the first way: (A-B-C)

We would have:

(A-D) — (B) — (C)
(A) — (B-D) —(C)
(A) — (B) — (CD)******

Case 2: this time, instead, for step 1 we end up choosing a group of 3 balls that is: (A—B—D) and 1 left over is C

Again there would be 3! = 6 different ways to arrange this group in the 3 places/boxes

A-B-D
A-D-B
etc.

The one left over is C — and we could arrange this in 3 ways for each one of the above 6 permutations. Let’s just choose 1 of the 6: (A—B—D)

We would have:

(A-C) — (B) — (D)
(A) — (B - C) — (D)
(A) — (B) — (D-C)****

However, look at the 2 starred ways:

The only difference is the ORDER in which we placed Balls D and C in the 3rd box. And since the order of the balls in each box does NOT MATTER, we have double counted the number of ways the distribution of balls to boxes can be done.

72 would be the correct answer if the ORDER MATTERED of the 2 balls that are placed together in the box

so you can “unorder” the 72 arrangements by dividing the 72 ways by 2! (Since for every 1 unique arrangement we want to keep we accidentally counted 2). This will get you to the correct answer of 36

or

From the beginning, in order to avoid accidentally ordering/arranging the 2 balls that end up in the box together, you can follow the below procedure.

Step 1: out of the 4 balls, first find how many ways we can choose an unordered group of 2 balls to be placed in a box together.

(4 c 2) = 6 ways

AND

Step 2; for each of these 6 ways, there are 3 choices of boxes in which you can place the unordered group of 2 balls:

(3 c 1) = 3 ways

AND

Step 3: finally, for the last 2 remaining balls, we can arrange these 2 balls in 2 boxes in:

2! = 2 ways

(4 c 2) * (3 c 1) * (2!) =

6 * 3 * 2 =

36 ways

C is the answer

Edit:
The answer above mine is also a clever way to get to the answer.

Step 1: You can figure out how many groups of 2 balls you can create out of 4. When you choose the 2 balls you want to include, you have automatically separated the other 2 balls from the group you’ve chosen.

(4 c 2) = 6 ways

for each of these 6 ways, we can assume the 2 unchosen balls are each part of their own separate group—— so that we would have 3 groupings of balls for each one of the 6 cases

These 6 cases would be grouped as follows:

(AB) - (C) - (D)

(AC) - (B) - (D)

(AD) - (B) - (C)

(A) - (BC) - (D)

(A) - (BD) - (C)

(A) - (B) - (CD)

AND

Step 2: for each of the 6 cases above, we have 3 boxes in which to arrange the 3 “distinct items”:

3! = 6 arrangements

(4 c 2) * (3!) =

(6) * (6) = 36 ways


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First ball can go in any of 3 boxes
Second ball can go in any of 2 remaining boxes
Third ball has to go in 1 remaining box
Fourth ball can go in any of 3 boxes

So, total = 3*2*1*3 = 18

KarishmaB, please let me know where I am going wrong.

Which balls to be paired up ?

4!/2!2! = 6

Now you have two balls and a pair, three objects.

Three objects can be arranged

3! = 6 ways

Total ways = 6×6 = 36


A visual approach:

First bucket has 4 choices of ball, 2nd 3 choices and third 2 choices equals

24 choices

Last ball has 3 choices, so

3×24 = 72.

However, this approach counts as distinct choices in the two ball bucket whether a given ball is first or second, which doesn't matter, so the total needs to be divided by 2

72/2=36

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