Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 1011:30 AM EST
-12:30 PM EST
Dreaming of getting into ISB but worried about your GMAT score? Bhavya’s journey shows that a strong profile can be just as important! Discover how she gained admission to the prestigious ISB with a 670 GMAT in the competitive R2, with GMATWhiz.
Nov 1005:30 AM PST
-07:30 AM PST
The Focus Edition allows barely 2 minutes per question. Hence, rereading portions of passages is not an option. Learn our Involved and Evolved reading strategy, the most efficacious solution to time saving on RC and score high on GMAT Verbal!
Nov 1305:30 AM PST
-06:00 PM PST
We are excited to bring you the Fall edition of the MBA Spotlight Fair! Meet admission officers from the top 20 MBA programs—including HBS, Stanford, and Wharton—as you prepare for Round 2 applications.
Nov 1308:00 PM PST
-09:00 PM PST
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Nov 1611:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.- Nov 17
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease.
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
40% (02:34) correct
60%
(02:39)
wrong
based on 542
sessions
History
Date
Time
Result
Not Attempted Yet
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
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16
29 Jul 2010, 19:42
Kudos
Bookmarks
dungtd
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!!!
(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: \(x=k+8=7+8=15\).
Answer: C (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.
29 Jul 2010, 18:41
Kudos
Bookmarks
bogos
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!
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
