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

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Intern
Joined: 13 Jun 2016
Posts: 13
Marshall has 4 different colored balls and 3 different boxes.  [#permalink]

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01 Jul 2018, 21:37
1
00:00

Difficulty:

85% (hard)

Question Stats:

30% (01:46) correct 70% (01:47) wrong based on 35 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.81
C.36
D.24
E.72
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3325
Location: India
GPA: 3.12
Re: Marshall has 4 different colored balls and 3 different boxes.  [#permalink]

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02 Jul 2018, 01:11
1
DarkBullRider 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.81
C.36
D.24
E.72

Since all the 3 boxes must have at least one ball, the three possibilities in which the four balls
A, B, C, and D can be arranged in 3 boxes are AB-C-D | C-AB-D | D-AB-C (3 ways)

The total ways of choosing 2 balls from 4 is $$C_2^4 = 6$$. There are 2 ways of choosing the
balls which goes into the second box and one way of choosing the ball which goes into the 3rd box.

Therefore, there are 3*6*2*1 = 36(Option C) ways of putting the 4 balls in the 3 boxes.
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Joined: 14 Jun 2018
Posts: 211
Re: Marshall has 4 different colored balls and 3 different boxes.  [#permalink]

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02 Jul 2018, 05:15
There are 4 balls and 3 boxes. (lets call them ball 1 , ball 2 etc)

Two of the balls must be in pair which can be done in 4c2 ways = 4!/2!2! = 6
Once you select the pair , remaining is selected by default

so now you have something like [1-2] [3] [4]. Now this can be arranged in 3! ways = 6 ways.

total ways = 6*6 = 36
Re: Marshall has 4 different colored balls and 3 different boxes. &nbs [#permalink] 02 Jul 2018, 05:15
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