GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 26 Jan 2020, 13:00

### GMAT Club Daily Prep

#### 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

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

# Mrs. Smith has been given film vouchers. Each voucher allows

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Joined: 09 Dec 2008
Posts: 22
Location: Vietnam
Schools: Somewhere
Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

### Show Tags

Updated on: 08 Jan 2013, 03:16
5
65
00:00

Difficulty:

95% (hard)

Question Stats:

37% (02:30) correct 63% (02:39) wrong based on 292 sessions

### HideShow timer Statistics

Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?

(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16

Originally posted by dungtd on 29 Jul 2010, 14:24.
Last edited by Bunuel on 08 Jan 2013, 03:16, edited 1 time in total.
RENAMED THE TOPIC.
Math Expert
Joined: 02 Sep 2009
Posts: 60646

### Show Tags

29 Jul 2010, 19:42
23
27
dungtd wrote:
Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16

Could someone help me out? Thanks a lot!!!

Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ be $$k$$.

In how many ways we can distribute $$k$$ identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.

Let $$k=5$$. And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).

Consider:

$$ttttt|||$$
We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:

$$ttttt|||$$
Means that first nephew will get all the tickets,

$$|t|ttt|t$$
Means that first got 0, second 1, third 3, and fourth 1

And so on.

How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$t$$'s and 3 $$|$$'s are identical, so $$\frac{8!}{5!3!}=56$$. Basically it's the number of ways we can pick 3 separators out of 5+3=8: $$8C3$$.

So, # of ways to distribute 5 tickets among 4 people is $$(5+4-1)C(4-1)=8C3$$.

For $$k$$ it will be the same: # of ways to distribute $$k$$ tickets among 4 persons (so that each can get from zero to $$k$$) would be $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$. --> $$k=7$$. Plus the 8 tickets we booked earlier: $$x=k+8=7+8=15$$.

P.S. How to solve $$(k+1)(k+2)(k+3)=3!*120=720$$: 720, three digit integer ending with 0, is the product of three consecutive integers. Obviously one of them must be multiple of 5: try $$5*6*7=210<720$$, next possible triplet $$8*9*10=720$$, OK. So $$k+1=8$$ --> $$k=7$$.

P.P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is $$n+r-1C_{r-1}$$.

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is $$n-1C_{r-1}$$.

Hope it helps.
_________________
Manager
Joined: 20 Mar 2010
Posts: 64

### Show Tags

29 Jul 2010, 18:41
20
3
bogos wrote:
It took me 10 minutes to solve it
First, it is remarked that all vouchers are equivalent.
Let's call nephews A, B, C, D and their number of vouchers (2+a), (2+b), (2+c), and (2+d), respectively, where a, b, c, d non negative integers (since each of them has at least 2 vouchers).
If n is the total number of vouchers, then:
a+b+c+d=n-8 or a+b+c+d=m with m=n-8
Let X=a+b and Y=c+d
so X+Y=m and (m+1) possible couples are (X=0, Y=m), (X=1, Y=m-1), ..., (X=m, Y=0)
On the other hand, (a, b) such that a+b=k and a,b>=0, the number of such couples is equal to k+1.
Therefore, generic solution is:
1*(m+1)+2*m+3*(m-1)+...+m*2+(m+1)*1
To get 120 different ways, m equals to 7, or n=15, thus C.

Sure this is not the easiest way to deal with this problem!

Actually there is a formula for this
The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items = n+r-1Cr-1.
Here if each nephew gets at least 2 tickets we have n-8 tickets left to distribute among 4 nephews and each one of them can receive 0,1,2..n-8 items.
Total no of ways that can happen = (n-8)+(4-1)C(4-1)
= n-5C3
n-5C3=120 for n=15 .So C
##### General Discussion
Intern
Joined: 03 Mar 2010
Posts: 34

### Show Tags

29 Jul 2010, 15:06
1
It took me 10 minutes to solve it
First, it is remarked that all vouchers are equivalent.
Let's call nephews A, B, C, D and their number of vouchers (2+a), (2+b), (2+c), and (2+d), respectively, where a, b, c, d non negative integers (since each of them has at least 2 vouchers).
If n is the total number of vouchers, then:
a+b+c+d=n-8 or a+b+c+d=m with m=n-8
Let X=a+b and Y=c+d
so X+Y=m and (m+1) possible couples are (X=0, Y=m), (X=1, Y=m-1), ..., (X=m, Y=0)
On the other hand, (a, b) such that a+b=k and a,b>=0, the number of such couples is equal to k+1.
Therefore, generic solution is:
1*(m+1)+2*m+3*(m-1)+...+m*2+(m+1)*1
To get 120 different ways, m equals to 7, or n=15, thus C.

Sure this is not the easiest way to deal with this problem!
Intern
Joined: 09 Dec 2008
Posts: 22
Location: Vietnam
Schools: Somewhere

### Show Tags

29 Jul 2010, 16:04
bogos wrote:
It took me 10 minutes to solve it
First, it is remarked that all vouchers are equivalent.
Let's call nephews A, B, C, D and their number of vouchers (2+a), (2+b), (2+c), and (2+d), respectively, where a, b, c, d non negative integers (since each of them has at least 2 vouchers).
If n is the total number of vouchers, then:
a+b+c+d=n-8 or a+b+c+d=m with m=n-8
Let X=a+b and Y=c+d
so X+Y=m and (m+1) possible couples are (X=0, Y=m), (X=1, Y=m-1), ..., (X=m, Y=0)
On the other hand, (a, b) such that a+b=k and a,b>=0, the number of such couples is equal to k+1.
Therefore, generic solution is:
1*(m+1)+2*m+3*(m-1)+...+m*2+(m+1)*1
To get 120 different ways, m equals to 7, or n=15, thus C.

Sure this is not the easiest way to deal with this problem!

Anyway, this is a good way to solve this tough problem! Thanks
Intern
Joined: 09 Dec 2008
Posts: 22
Location: Vietnam
Schools: Somewhere

### Show Tags

29 Jul 2010, 19:57
2
Bunuel wrote:
dungtd wrote:
Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16

Could someone help me out? Thanks a lot!!!

Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ be $$k$$.

In how many ways we can distribute $$k$$ identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.

Let $$k=5$$. And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).

Consider:

$$ttttt|||$$
We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:

$$ttttt|||$$
Means that first nephew will get all the tickets,

$$|t|ttt|t$$
Means that first got 0, second 1, third 3, and fourth 1

And so on.

How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$t$$'s and 3 $$|$$'s are identical, so $$\frac{8!}{5!3!}=56$$. Basically it's the number of ways we can pick 3 separators out of 5+3=8: $$8C3$$.

So, # of ways to distribute 5 tickets among 4 people is $$(5+4-1)C(4-1)=8C3$$.

For $$k$$ it will be the same: # of ways to distribute $$k$$ tickets among 4 persons (so that each can get from zero to $$k$$) would be $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$. --> $$k=7$$. Plus the 8 tickets we booked earlier: $$x=k+8=7+8=15$$.

P.S. How to solve $$(k+1)(k+2)(k+3)=3!*120=720$$: 720, three digit integer ending with 0, is the product of three consecutive integers. Obviously one of them must be multiple of 5: try $$5*6*7=210<720$$, next possible triplet $$8*9*10=720$$, OK. So $$k+1=8$$ --> $$k=7$$.

P.P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is $$n+r-1C_{r-1}$$.

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is $$n-1C_{r-1}$$.

Hope it helps.

Thanks for your great explanation!
Manager
Joined: 20 Dec 2010
Posts: 156
Schools: UNC Duke Kellogg

### Show Tags

02 Jul 2011, 18:53
5 stars for the "stars and bars" method! The coolest trick ever!
Manager
Joined: 11 Apr 2011
Posts: 71

### Show Tags

06 Jul 2011, 11:22
really great trick!!
Intern
Joined: 28 Mar 2011
Posts: 23

### Show Tags

23 Jul 2011, 23:52
Hi Bunuel,

Why do we assume k=5?

What would happen if we assume some other value of k?

Thanks!
Intern
Joined: 13 Oct 2012
Posts: 37
Concentration: General Management, Leadership
Schools: IE '15 (A)
GMAT 1: 760 Q49 V46

### Show Tags

07 Jan 2013, 12:04
this formula is worth remembering:-
The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items = n+r-1Cr-1.

Thanks a lot
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10011
Location: Pune, India

### Show Tags

07 Jan 2013, 21:10
10
2
dungtd wrote:
Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16

Could someone help me out? Thanks a lot!!!

This question is a spin on 'distribute n identical things among m people (some people may get none)'

Since she must give each of the 4 nephews atleast 2 vouchers, she must have atleast 8. But she can split the vouchers in 120 ways so she certainly has more than 8. Let's assume she gives 2 to each nephew and has the leftover vouchers in her hand. This becomes your standard 'distribute n identical things among m people (some people may get none)' question.

Assume the answer is 15 (it's in the middle) and she has 15 - 8 = 7 vouchers left in her hand. In how many ways can she distribute them among 4 people?

In 10!/7!*3! = 10*9*8/(3*2) = 120 ways.
This is the required answer so (C)

This concept is explained in detail here: http://www.veritasprep.com/blog/2011/12 ... 93-part-1/

Check method II or question 2 on the link above.

Note: Say, you had obtained a number less than 120. You would have assumed 16 next and repeated the calculation. Had you obtained a number more than 120, you would have assumed 13 or 14 and done the calculation. 2 iterations would have certainly given you the answer.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern
Joined: 07 Sep 2014
Posts: 17
Location: United States (MA)
Concentration: Finance, Economics
Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

### Show Tags

20 Oct 2014, 13:01
Bunuel wrote:
dungtd wrote:
Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16

Could someone help me out? Thanks a lot!!!

Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ be $$k$$.

In how many ways we can distribute $$k$$ identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.

Let $$k=5$$. And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).

Consider:

$$ttttt|||$$
We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:

$$ttttt|||$$
Means that first nephew will get all the tickets,

$$|t|ttt|t$$
Means that first got 0, second 1, third 3, and fourth 1

And so on.

How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$t$$'s and 3 $$|$$'s are identical, so $$\frac{8!}{5!3!}=56$$. Basically it's the number of ways we can pick 3 separators out of 5+3=8: $$8C3$$.

So, # of ways to distribute 5 tickets among 4 people is $$(5+4-1)C(4-1)=8C3$$.

For $$k$$ it will be the same: # of ways to distribute $$k$$ tickets among 4 persons (so that each can get from zero to $$k$$) would be $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$. --> $$k=7$$. Plus the 8 tickets we booked earlier: $$x=k+8=7+8=15$$.

P.S. How to solve $$(k+1)(k+2)(k+3)=3!*120=720$$: 720, three digit integer ending with 0, is the product of three consecutive integers. Obviously one of them must be multiple of 5: try $$5*6*7=210<720$$, next possible triplet $$8*9*10=720$$, OK. So $$k+1=8$$ --> $$k=7$$.

P.P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is $$n+r-1C_{r-1}$$.

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is $$n-1C_{r-1}$$.

Hope it helps.

Bunuel,

Blast from the past, but I'm just a bit unclear about the rearrangement here, and why/how the k! cancels out when we multiply by the 3!

For $$k$$ it will be the same: # of ways to distribute $$k$$ tickets among 4 persons (so that each can get from zero to $$k$$) would be $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$
Math Expert
Joined: 02 Sep 2009
Posts: 60646
Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

### Show Tags

20 Oct 2014, 14:10
2
bsmith37 wrote:
Bunuel wrote:
dungtd wrote:
Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16

Could someone help me out? Thanks a lot!!!

Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ be $$k$$.

In how many ways we can distribute $$k$$ identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.

Let $$k=5$$. And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).

Consider:

$$ttttt|||$$
We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:

$$ttttt|||$$
Means that first nephew will get all the tickets,

$$|t|ttt|t$$
Means that first got 0, second 1, third 3, and fourth 1

And so on.

How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$t$$'s and 3 $$|$$'s are identical, so $$\frac{8!}{5!3!}=56$$. Basically it's the number of ways we can pick 3 separators out of 5+3=8: $$8C3$$.

So, # of ways to distribute 5 tickets among 4 people is $$(5+4-1)C(4-1)=8C3$$.

For $$k$$ it will be the same: # of ways to distribute $$k$$ tickets among 4 persons (so that each can get from zero to $$k$$) would be $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$. --> $$k=7$$. Plus the 8 tickets we booked earlier: $$x=k+8=7+8=15$$.

P.S. How to solve $$(k+1)(k+2)(k+3)=3!*120=720$$: 720, three digit integer ending with 0, is the product of three consecutive integers. Obviously one of them must be multiple of 5: try $$5*6*7=210<720$$, next possible triplet $$8*9*10=720$$, OK. So $$k+1=8$$ --> $$k=7$$.

P.P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is $$n+r-1C_{r-1}$$.

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is $$n-1C_{r-1}$$.

Hope it helps.

Bunuel,

Blast from the past, but I'm just a bit unclear about the rearrangement here, and why/how the k! cancels out when we multiply by the 3!

For $$k$$ it will be the same: # of ways to distribute $$k$$ tickets among 4 persons (so that each can get from zero to $$k$$) would be $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$

(k+3)! = k!*(k+1)(k+2)(k+3), so $$\frac{(k+3)!}{k!3!}=\frac{k!*(k+1)(k+2)(k+3)}{k!3!}=\frac{(k+1)(k+2)(k+3)}{3!}$$.
_________________
Intern
Joined: 07 Sep 2014
Posts: 17
Location: United States (MA)
Concentration: Finance, Economics
Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

### Show Tags

20 Oct 2014, 15:00
Quote:

(k+3)! = k!*(k+1)(k+2)(k+3), so $$\frac{(k+3)!}{k!3!}=\frac{k!*(k+1)(k+2)(k+3)}{k!3!}=\frac{(k+1)(k+2)(k+3)}{3!}$$.

that is what i figured. thank you.
Director
Joined: 17 Dec 2012
Posts: 622
Location: India
Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

### Show Tags

26 Oct 2015, 22:51
1
Let us assume the total number of vouchers is 15. Each nephew should get at least 2 vouchers. The distribution can contain three 2 vouchers, two 2 vouchers , one two voucher or no 2 voucher. For three 2 vouchers the possibilities are 2 2 2 9, 2 2 9 2, 2 9 2 2 and 9 2 2 2 . So there are 4 possibilities. For two 2 vouchers, one 2 voucher and no two voucher, the possibilities can be shown to be 36, 60 and 20. So the total number of possibilities are 120 as required.

Even if we did not choose 15 as the number of vouchers we can arrive at it with a couple of choices.
_________________
Srinivasan Vaidyaraman
Magical Logicians

Holistic and Holy Approach
Intern
Joined: 27 Sep 2015
Posts: 15
Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

### Show Tags

26 Dec 2015, 09:45
I was wondering does the question mean
1. each nephew gets 2 vouchers first,
2. the remaining vouchers can be distributed in 120 ways.

so each gets 2 first. 2*4=8
then the remaining vouchers can be distributed in 120 ways
(K+3)! / K!*3! = 120
(K+3) (K+2) (K+1) = 720
plug in numbers K = 7
7+ the original 8 = 15.
is there anything wrong with my understanding ?
thanks
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10011
Location: Pune, India
Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

### Show Tags

29 Dec 2015, 02:41
1
VeritasPrepKarishma wrote:
dungtd wrote:
Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16

Could someone help me out? Thanks a lot!!!

This question is a spin on 'distribute n identical things among m people (some people may get none)'

Since she must give each of the 4 nephews atleast 2 vouchers, she must have atleast 8. But she can split the vouchers in 120 ways so she certainly has more than 8. Let's assume she gives 2 to each nephew and has the leftover vouchers in her hand. This becomes your standard 'distribute n identical things among m people (some people may get none)' question.

Assume the answer is 15 (it's in the middle) and she has 15 - 8 = 7 vouchers left in her hand. In how many ways can she distribute them among 4 people?

In 10!/7!*3! = 10*9*8/(3*2) = 120 ways.
This is the required answer so (C)

This concept is explained in detail here: http://www.veritasprep.com/blog/2011/12 ... 93-part-1/

Check method II or question 2 on the link above.

Note: Say, you had obtained a number less than 120. You would have assumed 16 next and repeated the calculation. Had you obtained a number more than 120, you would have assumed 13 or 14 and done the calculation. 2 iterations would have certainly given you the answer.

Quote:
I have one confusion though. So, when we distribute the remaining 7 vouchers we are dividing by 7! and 3!. These take care of the arrangements with 7 identical vouchers and 3 identical bars (vvvvvvv|||), right?. But what about arrangements like vvv|v|vv|v? Here, how do we take care of the identical arrangements with vvv, vv? If we do not take care of them, then with vvv, there will be additional 3! = 6 arrangements multiplied by 2! arranments for vv, right?

Can you please clarify? I have this confusion for other similar porblems. Maybe I am missing some point here.

Consider this question: How many words can you make using all letters of GOSSIP? This is 6!/2!

How about number of words using all letters of MMIISS? This is 6!/2!*2!*2! (this includes all words such as MIISMS, ISMISM, ISSMMI etc)

On the same lines, consider VVVVVVVIII - Is it any different from above? This would be 10!/7!*3! (this would include all words such as VVVIVIVVIV)
We have already taken care of all identical Vs when we divide by 7!
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern
Joined: 19 Oct 2014
Posts: 35
Location: India
Concentration: Finance, Entrepreneurship
GMAT 1: 600 Q48 V25
GMAT 2: 670 Q49 V31
GPA: 3.26
WE: Operations (Manufacturing)
Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

### Show Tags

24 Apr 2018, 01:37
Bunuel wrote:

P.P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is $$n+r-1C_{r-1}$$.

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is $$n-1C_{r-1}$$.

Hope it helps.

Hi Bunuel,
Can you help explain the formula(or share the link if it's already done) for finding the number of ways of dividing n identical items among r persons, when each one of whom receives at least one item ?
In this case, how does the concept of separator work. To me, it suggests the formula will have to be $$n+rC_{r}$$. Can you help me please understand this.
Intern
Joined: 28 Jul 2018
Posts: 11
Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

### Show Tags

25 Feb 2019, 04:33
Bunuel wrote:
dungtd wrote:
Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16

Could someone help me out? Thanks a lot!!!

Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ be $$k$$.

In how many ways we can distribute $$k$$ identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.

Let $$k=5$$. And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).

Consider:

$$ttttt|||$$
We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:

$$ttttt|||$$
Means that first nephew will get all the tickets,

$$|t|ttt|t$$
Means that first got 0, second 1, third 3, and fourth 1

And so on.

How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$t$$'s and 3 $$|$$'s are identical, so $$\frac{8!}{5!3!}=56$$. Basically it's the number of ways we can pick 3 separators out of 5+3=8: $$8C3$$.

So, # of ways to distribute 5 tickets among 4 people is $$(5+4-1)C(4-1)=8C3$$.

For $$k$$ it will be the same: # of ways to distribute $$k$$ tickets among 4 persons (so that each can get from zero to $$k$$) would be $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$. --> $$k=7$$. Plus the 8 tickets we booked earlier: $$x=k+8=7+8=15$$.

P.S. How to solve $$(k+1)(k+2)(k+3)=3!*120=720$$: 720, three digit integer ending with 0, is the product of three consecutive integers. Obviously one of them must be multiple of 5: try $$5*6*7=210<720$$, next possible triplet $$8*9*10=720$$, OK. So $$k+1=8$$ --> $$k=7$$.

P.P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is $$n+r-1C_{r-1}$$.

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is $$n-1C_{r-1}$$.

Hope it helps.

Bunuel

Hi,

let's assume that Mr. Smith has 15 vouchers and 4 nephews whom she want to give the vouchers with the restriction that everyone gets at least 2 vouchers.

No we should use the formula ("The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is $$n-1C_{r-1}$$"

This would be $$15-2C_{4-1}$$. This does not result in 120. Where am I wrong?
Manager
Joined: 23 Aug 2017
Posts: 117
Schools: ISB '21 (A)
Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

### Show Tags

25 Feb 2019, 07:35
1
Why dont we consider the no of ways in which 8 vouchers are to be distributed among the 4 nephews equally? Isnt that also included within 120 ways?

Re: Mrs. Smith has been given film vouchers. Each voucher allows   [#permalink] 25 Feb 2019, 07:35

Go to page    1   2    Next  [ 28 posts ]

Display posts from previous: Sort by

# Mrs. Smith has been given film vouchers. Each voucher allows

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne