Five Distinct balls needs to be placed in three boxes

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Joepc: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Tue Sep 06, 2016 11:38 pm

by Joepc » Wed Feb 22, 2017 9:12 pm

Question :- Five balls needs to be placed in three boxes. Each box can hold all the five balls. In how many ways can the balls be placed in the boxes if no box can be empty and all balls and boxes are different

Please Help to solve this type of division and distribution combination sums

I approached like below

Case A :- 1,1,3

From 5 distinct balls i can group them into (1,1,3) in 5!/3!*2! no of ways
After grouping each group can be put to 3! ways
So together it is (5!/3!*2!)*3!
= 60 ways

Case B :- 1,1 2

From 5 distinct balls i can group them into (1,1 2) in 5!/2!*2! no of ways
After grouping each group can be put to 3! ways
So together it is (5!/2!*2!)*3!

= 180 ways

Case A + Case B
= 60 + 180 = 240 ways

Is my approach correct?

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Source: Beat The GMAT — Problem Solving | Wed Feb 22, 2017 9:12 pm

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu Feb 23, 2017 3:27 am

Five balls of different colors are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many different ways can we we place the balls so that no box remains empty?

A. 150 B. 10 C. 60 D. 300 E. 375

Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marble
Number of box options for the box with 3 marbles = 3. (Any of the 3 boxes.)
For this box, the number of ways to choose 3 marbles from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of marble options for the next box = 2. (Either of the 2 remaining marbles.)
Number of marble options for the last box = 1. (Only 1 marble left.)
To combine these options, we multiply:
3*10*2*1 = 60.

Case 2: 1 box has 1 marble, the other 2 boxes each have 2 marbles
Number of box options for the box with 1 marble = 3. (Any of the 3 boxes.)
Number of marbles that could be placed in this box = 5. (Any of the 5 marbles.)
From the 4 remaining marbles, the number of ways to choose 2 marbles for the next box = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining marbles, the number of ways to choose 2 marbles for the last box = 2C2 = (2*1)/(2*1) = 1.
To combine these options, we multiply:
3*5*6*1 = 90.

Total ways = 60+90 = 150.

The correct answer is A.

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Joepc: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Tue Sep 06, 2016 11:38 pm

by Joepc » Thu Feb 23, 2017 12:36 pm

GMATGuruNY wrote:
Five balls of different colors are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many different ways can we we place the balls so that no box remains empty?

A. 150 B. 10 C. 60 D. 300 E. 375
Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marble
Number of box options for the box with 3 marbles = 3. (Any of the 3 boxes.)
For this box, the number of ways to choose 3 marbles from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of marble options for the next box = 2. (Either of the 2 remaining marbles.)
Number of marble options for the last box = 1. (Only 1 marble left.)
To combine these options, we multiply:
3*10*2*1 = 60.

Case 2: 1 box has 1 marble, the other 2 boxes each have 2 marbles
Number of box options for the box with 1 marble = 3. (Any of the 3 boxes.)
Number of marbles that could be placed in this box = 5. (Any of the 5 marbles.)
From the 4 remaining marbles, the number of ways to choose 2 marbles for the next box = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining marbles, the number of ways to choose 2 marbles for the last box = 2C2 = (2*1)/(2*1) = 1.
To combine these options, we multiply:
3*5*6*1 = 90.

Total ways = 60+90 = 150.

The correct answer is A.

Thanks a Ton Thus solved My confusion

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Thu Mar 02, 2017 5:08 pm

This isn't really a GMAT topic (marbles in boxes), but if you're curious, there are a few easy formulas that address the basic cases.

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