Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In how many ways can 5 different candiesbe distributed among [#permalink]

Show Tags

21 Oct 2012, 17:52

9

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

47% (02:09) correct
53% (01:14) wrong based on 148 sessions

HideShow timer Statistics

In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy) (A) 4^5 (B) 5^4 (C) 5! (D) 4! (E) 4!*5!

Re: In how many ways can 5 different candiesbe distributed among [#permalink]

Show Tags

22 Oct 2012, 08:29

2

This post received KUDOS

voodoochild wrote:

In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy) (A) 4^5 (B) 5^4 (C) 5! (D) 4! (E) 4!*5!

If I add these numbers, it doesn't equal to 1024. What's my mistake?

Thanks

5-N-N-N : 4!/3! * 5C5 = 4 - OK 4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60 - OK, divide by 2! because the two zero's (N) are indistinguishable 3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120 - OK 3-1-1-N : 4!/2! * 5C3*2C2 = 12*10 = 100 - it should be 4! * 5C3 = 24*10 = 240, you should not divide by 2!, each ball is distinct 2-2-1-N : 4!/2! * 5C2*3C2*1C1 *(1/2) (1/2 because we are double counting 2-2) = 12 * 15 = 180 - NO - it should be 4! * 5C2 * 3C2 * 1C1/2! = 360, doesn't matter which group of 2 you have chosen first, but later, when permuting them, it counts 2-1-1-1 : 4!/3! * 5C2*3C3 = 4*10=40 - NO - it should be 4! * 5C2 = 240, all the balls are distinct, so you should not divide by 3!.

The total is 4 + 60 + 120 + 240 + 360 + 240 = 1024.

But why sweat all the way along????
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: In how many ways can 5 different candiesbe distributed among [#permalink]

Show Tags

31 Oct 2012, 12:16

EvaJager wrote:

voodoochild wrote:

In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy) (A) 4^5 (B) 5^4 (C) 5! (D) 4! (E) 4!*5!

If I add these numbers, it doesn't equal to 1024. What's my mistake?

Thanks

5-N-N-N : 4!/3! * 5C5 = 4 - OK 4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60 - OK, divide by 2! because the two zero's (N) are indistinguishable 3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120 - OK 3-1-1-N : 4!/2! * 5C3*2C2 = 12*10 = 100 - it should be 4! * 5C3 = 24*10 = 240, you should not divide by 2!, each ball is distinct 2-2-1-N : 4!/2! * 5C2*3C2*1C1 *(1/2) (1/2 because we are double counting 2-2) = 12 * 15 = 180 - NO - it should be 4! * 5C2 * 3C2 * 1C1/2! = 360, doesn't matter which group of 2 you have chosen first, but later, when permuting them, it counts 2-1-1-1 : 4!/3! * 5C2*3C3 = 4*10=40 - NO - it should be 4! * 5C2 = 240, all the balls are distinct, so you should not divide by 3!.

The total is 4 + 60 + 120 + 240 + 360 + 240 = 1024.

But why sweat all the way along????

Why do you have to divide by 2! on 2-2-1-N but don't have to divide by 2! on 3-1-1-N?

Re: In how many ways can 5 different candiesbe distributed among [#permalink]

Show Tags

31 Oct 2012, 12:48

1

This post received KUDOS

BN1989 wrote:

EvaJager wrote:

voodoochild wrote:

In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy) (A) 4^5 (B) 5^4 (C) 5! (D) 4! (E) 4!*5!

If I add these numbers, it doesn't equal to 1024. What's my mistake?

Thanks

5-N-N-N : 4!/3! * 5C5 = 4 - OK 4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60 - OK, divide by 2! because the two zero's (N) are indistinguishable 3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120 - OK 3-1-1-N : 4!/2! * 5C3*2C2 = 12*10 = 100 - it should be 4! * 5C3 = 24*10 = 240, you should not divide by 2!, each ball is distinct 2-2-1-N : 4!/2! * 5C2*3C2*1C1 *(1/2) (1/2 because we are double counting 2-2) = 12 * 15 = 180 - NO - it should be 4! * 5C2 * 3C2 * 1C1/2! = 360, doesn't matter which group of 2 you have chosen first, but later, when permuting them, it counts 2-1-1-1 : 4!/3! * 5C2*3C3 = 4*10=40 - NO - it should be 4! * 5C2 = 240, all the balls are distinct, so you should not divide by 3!.

The total is 4 + 60 + 120 + 240 + 360 + 240 = 1024.

But why sweat all the way along????

Why do you have to divide by 2! on 2-2-1-N but don't have to divide by 2! on 3-1-1-N?

In the case of 2-2-1-N we divide by 2! because we choose 2 groups of 2 one after the other and order when choosing them doesn't matter. Anyway, we permute them latter (see the factor of 4!). Similarly, in the case of 3-1-1-N we should divide by 2! if we choose one single ball after the other 4! * 5C3 * 2C1 * 1C1/2!. If it is clear that we split the two remaining balls after choosing 3, we don't need the 2C1 and 1C1 factors. That's why 4! * 5C3.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: In how many ways can 5 different candiesbe distributed among [#permalink]

Show Tags

01 Nov 2012, 09:08

Thanks for your reply Eva, but I still haven’t fully understood it yet. Let’s use the 3-1-1-N example, let’s say the candies are named A, B, C, D and E: One possible outcome would be ABC-D-E-N, for this outcome there are 4! possible ways to distribute this outcome among the four people and 5C3 ways for the one person to get 3 candies so there are 240 possible arrangements of this kind. Now I don’t understand exactly how this would look like if we choose one candy after the other, so that this formula applies: 4! * 5C3 * 2C1 * 1C1/2! If you draw again ABC for the person who gets 3, you have D and E left for the next person so two choices and then you have again two choices about who out of the remaining two people gets the last candy?! So after ABC was picked you have 2C1*2!=4 options and you could permute this outcome in 4! ways, which obviously is wrong. How exactly does the process look like if I draw according to 4! * 5C3 * 2C1 * 1C1/2!?

Re: In how many ways can 5 different candiesbe distributed among [#permalink]

Show Tags

01 Nov 2012, 10:03

BN1989 wrote:

Thanks for your reply Eva, but I still haven’t fully understood it yet. Let’s use the 3-1-1-N example, let’s say the candies are named A, B, C, D and E: One possible outcome would be ABC-D-E-N, for this outcome there are 4! possible ways to distribute this outcome among the four people and 5C3 ways for the one person to get 3 candies so there are 240 possible arrangements of this kind. Now I don’t understand exactly how this would look like if we choose one candy after the other, so that this formula applies: 4! * 5C3 * 2C1 * 1C1/2! If you draw again ABC for the person who gets 3, you have D and E left for the next person so two choices and then you have again two choices about who out of the remaining two people gets the last candy?! So after ABC was picked you have 2C1*2!=4 options and you could permute this outcome in 4! ways, which obviously is wrong. How exactly does the process look like if I draw according to 4! * 5C3 * 2C1 * 1C1/2!?

How exactly does the process look like if I draw according to 4! * 5C3 * 2C1 * 1C1/2!?

Think of carrying out the process in two steps: 1) Divide the 5 candies into 3 groups - 3, 1, 1 Decide on the 3 candies somebody will get them - 5C3 - for example A, B, C (but think of all the possibilities) Which one is the first 1 single candy - 2C1 - D or E Which one is the second 1 single candy - 1C1 - E or D This would give 5C3 * 2C1 * 1C1 Divide by 2! because we are going to permute the groups of 3, 1, 1, 0, so we don't care about which single candy was chosen first. ABC, D, E is the same as ABC, E, D - what matters is that A, B, C are together, and D and E are single 2) After we have our groups of 3, 1, 1, 0 candies, we consider them as 4 distinct objects and just permute them - 4! - because we have 4 different kids.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

I am getting 8!/(3!*5!)=56. I am using the stars bars method. (n+k-1)C(k-1). Why is this answer incorrect?

There are various methods of solving P&C questions and there are many formulas. You need to know the exact situation in which you can use a particular formula or better yet, you should understand the reason a particular method works in a particular situation.

Stars bars method (or the formula (n+k-1)C(k-1)) is useful only when the objects to be distributed are identical. You split identical objects into different groups. You divide by 5! because the objects are identical. Here you can not use that method because the candies are different.

Check out these posts on when to use which method:

Re: In how many ways can 5 different candiesbe distributed among [#permalink]

Show Tags

03 Sep 2013, 00:17

5

This post received KUDOS

there are total 5 candies. Lets say A,B,C,D and E. Candy A can be distributed among any 4 children. So there are 4 ways of distributing candy A. similarly candy B,C,D and E can be distributed in 4 ways.

There are various methods of solving P&C questions and there are many formulas. You need to know the exact situation in which you can use a particular formula or better yet, you should understand the reason a particular method works in a particular situation.

Stars bars method (or the formula (n+k-1)C(k-1)) is useful only when the objects to be distributed are identical. You split identical objects into different groups. You divide by 5! because the objects are identical. Here you can not use that method because the candies are different.

Check out these posts on when to use which method:

Why is the same formula (n+r-1)C(r-1) used for number of ways of dividing n identical items among r persons whom can receive 0,1,2 or more items . Is the formula used even if each person can receive 2 or 3 or any item? Why isn't that being factored in to the formula?

How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 1 mango?

How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 2 mango?

How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 3 mango?

Are you saying this formula gives us the same answer for all these questions? (10+4-1)C(4-1)?

Certainly not. You will not have the same answer in every case. You can use the same formula but the value of n will vary in each case.

Say if every kid must receive at least one mango, you take any 4 mangoes from the bunch of 10 (they are identical) and give one to each of the 4 kids. You cannot do this in more than 1 way because all the mangoes are the same and all the kids are receiving exactly one mango. Next, you are left with 6 mangoes and 4 kids and you need to distribute these 6 among the kids such that a kid may get no mango and another may get all etc. (You can do this because all the kids have already got a mango each and hence our condition is already satisfied.) This boils down to our previous question except that the value of n = 6 now, not n = 10.

Now you use the formula as (6 + 4 - 1)C(4 - 1) = 9!/6!*3!

You can do the same thing for at least 2 mangoes too.
_________________

Re: In how many ways can 5 different candiesbe distributed among [#permalink]

Show Tags

12 Nov 2014, 15:38

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

In how many ways can 5 different candiesbe distributed among [#permalink]

Show Tags

09 Aug 2015, 15:14

Why do we ignore the case in which no candy is given to any child? It is literally in the question statement that the children could get zero candies:

Quote:

In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy)

Admittedly, I'm not a native English speaker but to me it's clear that I do not have to give a candy to any child. I can keep all candies to myself.

Nor does the question state that all five candies must be given out to children. I could give out 0, 1, 2, 3, 4, or all five.

So, using the awesome approach posted by shrikar23, we would get 5^5 because there are 5 ways of giving out each distinct candy: give it to no child, or to child 1, or to child 2 and so forth.

Why do we ignore the case in which no candy is given to any child? It is literally in the question statement that the children could get zero candies:

Quote:

In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy)

Admittedly, I'm not a native English speaker but to me it's clear that I do not have to give a candy to any child. I can keep all candies to myself.

Nor does the question state that all five candies must be given out to children. I could give out 0, 1, 2, 3, 4, or all five.

So, using the awesome approach posted by shrikar23, we would get 5^5 because there are 5 ways of giving out each distinct candy: give it to no child, or to child 1, or to child 2 and so forth.

The question states that you need to "distribute 5 candies". So all 5 candies must be given out. It is possible that one or more children may not get any candy and one or more may get more than one candies but all 5 have to be distributed.
_________________

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...