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

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Math Expert
Joined: 02 Sep 2009
Posts: 59720
Marshall has 4 different colored balls and 3 different boxes. In how  [#permalink]

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12 Nov 2019, 03:59
00:00

Difficulty:

95% (hard)

Question Stats:

16% (01:49) correct 84% (01:26) wrong based on 45 sessions

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Marshall has 4 different colored balls and 3 different boxes. In how many ways can he put these balls in the boxes so that no box is left empty and no ball is left out?

A. 18
B. 24
C. 36
D. 72
E. 81

Are You Up For the Challenge: 700 Level Questions

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Intern
Joined: 12 May 2019
Posts: 9
Location: India
Concentration: Marketing, General Management
Marshall has 4 different colored balls and 3 different boxes. In how  [#permalink]

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12 Nov 2019, 04:22
As per my caclulation, which i have done taking each case, the answer comes out to be 81

Any fast approach?
Manager
Joined: 10 Dec 2017
Posts: 151
Location: India
Re: Marshall has 4 different colored balls and 3 different boxes. In how  [#permalink]

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12 Nov 2019, 05:28
1
Bunuel wrote:
Marshall has 4 different colored balls and 3 different boxes. In how many ways can he put these balls in the boxes so that no box is left empty and no ball is left out?

A. 18
B. 24
C. 36
D. 72
E. 81

Are You Up For the Challenge: 700 Level Questions

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:)
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Re: Marshall has 4 different colored balls and 3 different boxes. In how  [#permalink]

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12 Nov 2019, 09:02
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

Bunuel wrote:
Marshall has 4 different colored balls and 3 different boxes. In how many ways can he put these balls in the boxes so that no box is left empty and no ball is left out?

A. 18
B. 24
C. 36
D. 72
E. 81

Are You Up For the Challenge: 700 Level Questions
SVP
Joined: 03 Jun 2019
Posts: 1884
Location: India
Re: Marshall has 4 different colored balls and 3 different boxes. In how  [#permalink]

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12 Nov 2019, 09:50
Bunuel wrote:
Marshall has 4 different colored balls and 3 different boxes. In how many ways can he put these balls in the boxes so that no box is left empty and no ball is left out?

A. 18
B. 24
C. 36
D. 72
E. 81

Are You Up For the Challenge: 700 Level Questions

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

IMO D
Intern
Joined: 08 Apr 2017
Posts: 2
Re: Marshall has 4 different colored balls and 3 different boxes. In how  [#permalink]

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16 Nov 2019, 07:50
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
Intern
Joined: 10 Sep 2019
Posts: 2
Re: Marshall has 4 different colored balls and 3 different boxes. In how  [#permalink]

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20 Nov 2019, 14:57
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.
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Marshall has 4 different colored balls and 3 different boxes. In how  [#permalink]

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20 Nov 2019, 18:55
Bunuel wrote:
Marshall has 4 different colored balls and 3 different boxes. In how many ways can he put these balls in the boxes so that no box is left empty and no ball is left out?

A. 18
B. 24
C. 36
D. 72
E. 81

Are You Up For the Challenge: 700 Level Questions

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.

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Re: Marshall has 4 different colored balls and 3 different boxes. In how  [#permalink]

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24 Nov 2019, 21:39
1

Solution

Given:
• Marshall has 4 different colored balls and 3 different boxes

To find:
• In how many ways can he put these balls in the boxes so that no box is left empty and no ball is left out

Approach and Working Out:
• The only possible case of splitting these four balls into three boxes = {1, 1, 2}
• So, only one box will have 2 balls in it and the rest two will have one each

Step 1: select that box in which two balls need to be placed
• Number of ways to do it = $$^3C_1 = 3$$

Step 2: select the two balls which need to be placed together in a box
Number of ways to do that = $$^4C_2 = 6$$

Step 3: the remaining two balls can be arranged in 2! = ways

Therefore, the required answer = 3 * 6 * 2 = 36 ways

Hence, the correct answer is Option C.

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Re: Marshall has 4 different colored balls and 3 different boxes. In how   [#permalink] 24 Nov 2019, 21:39
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