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# NEW SET of good PS(3)

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Re: NEW SET of good PS(3) [#permalink]

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28 May 2014, 11:48
Bunuel wrote:
5. 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

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. 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 please help solve this problem directly using the formulas provided?

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}.

Regards
Harish
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Re: NEW SET of good PS(3) [#permalink]

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11 Jul 2014, 22:29
Augustus wrote:
yangsta8 wrote:
Bunuel wrote:
1. ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A,B,C,D,E and F?
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

6 points in total to make triangles. I think a combination of any 3 will make a unique triangle so:
6C3 = 20

Will anyone please explain what the C means in the notation?

I guess, 6C3 stands for a Combination formula, i.e. we have 3 choices out of 6 without restrictions. This means 6!/3!*3!=20
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Re: NEW SET of good PS(3) [#permalink]

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09 Aug 2014, 06:06
Bunuel wrote:

As most of the problems was solved correctly, I'm posting only OAs. Please let me know if anyone needs any clarification.

1. ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A,B,C,D,E and F?
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

2. The function f is defined for all positive integers n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. If p is prime, then f(p) =
(A) P-1
(B) P-2
(C) (P+1)/2
(D) (P-1)/2
(E) 2

3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72

4.A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
(A) 20
(B) 36
(C) 48
(D) 60
(E) 84

5. 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

6. This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal to half the amount that he spends this year?
(A) 1/(r+2)
(B) 1/(2r+2)
(C) 1/(3r+2)
(D) 1/(r+3)
(E) 1/(2r+3)

7. Before being simplified, the instructions for computing income tax in Country Rwere to add 2 percent of one's annual income to the average(arithmetic mean)of 100units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax in Country R's currency, for a person in that country whose annual income is I?
(A) 50+I/200
(B) 50+3I/100
(C) 50+I/40
(D) 100+I/50
(E) 100+3I/100

8. How many positive integers less than 10,000 are such that the product of their digits is 210?
(A) 24
(B) 30
(C) 48
(D) 54
(E) 72

9. Find the number of selections that can be made taking 4 letters from the word"ENTRANCE".
(A) 70
(B) 36
(C) 35
(D) 72
(E) 32

Find in the above word, the number of arrangements using the 4 letters.

10. How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3?
(A) 72
(B) 76
(C) 78
(D) 80
(E) 84

Hello Bunuel, Could you please explain the approaches to solve Q9 and Q10.
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01 Sep 2015, 23:59
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05 Oct 2015, 16:33
2. The function f is defined for all positive integers n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. If p is prime, then f(p) =
(A) P-1
(B) P-2
(C) (P+1)/2
(D) (P-1)/2
(E) 2

Regarding the question 2, I think it should be choice B and not A
I agree to the way of solving here . BUT question says f(n) includes all integers LESS than N and does not say including N . Thus if F is a prime, shouldnt it be n-1(as n is not included) and another -1 (as 1 is a common factor, Thus n-2.

Apologies if I am wrong. PLease correct me .
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06 Oct 2015, 03:52
shreyashid wrote:
2. The function f is defined for all positive integers n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. If p is prime, then f(p) =
(A) P-1
(B) P-2
(C) (P+1)/2
(D) (P-1)/2
(E) 2

Regarding the question 2, I think it should be choice B and not A
I agree to the way of solving here . BUT question says f(n) includes all integers LESS than N and does not say including N . Thus if F is a prime, shouldnt it be n-1(as n is not included) and another -1 (as 1 is a common factor, Thus n-2.

Apologies if I am wrong. PLease correct me .

That's not correct. There is no reason whatsoever to exclude 1: 1 also does not have have a common factor with say 7 other than 1. So, you should include 1 as well.
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Re: NEW SET of good PS(3) [#permalink]

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10 Feb 2016, 01:39
I think your answer for the 8th question is wrong coz ur missing the cases when it is 5, 6 ,7 and 1 which are 24 more cases. So the answer should be 54 and not 30

For question 5 assume first that Mrs. Smith has given 8 tickets to her grandsons by giving 2 to each and has "x" tickets left. So she can now distribute these x tickets to her 4 grandsons in (x+3)C3 ways. This is selection without arrangement so we use this formula.
We thus get this to be equal to 120. Thus we get x+3 = 10 and x = 7
hence total tickets is 15

for the second question I think we should count 1, as there is no reason for not counting it and hence the answer should be p-1 only.[/quote]

Can anyone please explain from where this (X+3) came? And how (x+3)C3=120 is equal to x+3 = 10?

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Re: NEW SET of good PS(3) [#permalink]

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10 Feb 2016, 02:00
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alishandhanani17 wrote:
I think your answer for the 8th question is wrong coz ur missing the cases when it is 5, 6 ,7 and 1 which are 24 more cases. So the answer should be 54 and not 30

For question 5 assume first that Mrs. Smith has given 8 tickets to her grandsons by giving 2 to each and has "x" tickets left. So she can now distribute these x tickets to her 4 grandsons in (x+3)C3 ways. This is selection without arrangement so we use this formula.
We thus get this to be equal to 120. Thus we get x+3 = 10 and x = 7
hence total tickets is 15

for the second question I think we should count 1, as there is no reason for not counting it and hence the answer should be p-1 only.

Can anyone please explain from where this (X+3) came? And how (x+3)C3=120 is equal to x+3 = 10?

Hi,
Can anyone please explain from where this (X+3) came? And how (x+3)C3=120 is equal to x+3 = 10?
(x+3)C3= (x+3)!/3!(x+3-3)!= (x+3)!/3!(x)!
= (x+3)(x+2)(x+1)x!/3!x!= (x+3)(x+2)(x+1)/3!
now this is equal to 120..
(x+3)(x+2)(x+1)/3!=120
(x+3)(x+2)(x+1)=720..
now 720=8*9*10=(x+3)(x+2)(x+1)..
thus x=7 and x+3=10
hope it helps
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04 Jun 2016, 23:53
Bunuel wrote:
h2polo wrote:
Bunuel,

Quality problems as usual... and also as usual they have kicked my butt.

Can you please post your solution to Question 9?

It is driving me nuts!

Thanks again!

Find the number of selections that can be made taking 4 letters from the word "ENTRANCE".
(A) 70
(B) 36
(C) 35
(D) 72
(E) 32

This problem is not the one you'll see on real GMAT. As combinatorics problems on GMAT are quite straightforward. But still it could be good for practice.

We have 8 letters from which 6 are unique.

Possible scenarios for 4 letter selection are:
A. All letters are different;
B. 2 N-s and other letters are different;
C. 2 E-s and other letters are different;
D. 2 N-s, 2 E-s.

Let's count them separately:
A. All letters are different, means that basically we are choosing 4 letters form 6 distinct letters: 6C4=15;
B. 2 N-s and other letters are different: 2C2(2 N-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;
C. 2 E-s and other letters are different: 2C2(2 E-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;;
D. 2 N-s, 2 E-s: 2C2*2C2=1.

15+10+10+1=36

Hi,

I fail to understand as to why have you considered multiple scenario's for the 4 letter selection.
I believe that the formula of nCr i.e. 8C4 would consider all of the same. Are you trying to eradicate the possibility of some duplicates in the scenario?
Could you kindly enumerate the same principle in case we have to select 3 out of 5 letters from AABCD by mentioning all the possible selections, so we can understand what is not to be considered?

Thanks
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Re: NEW SET of good PS(3) [#permalink]

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05 Jun 2016, 00:03
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nishi999 wrote:
Bunuel wrote:
h2polo wrote:
Bunuel,

Quality problems as usual... and also as usual they have kicked my butt.

Can you please post your solution to Question 9?

It is driving me nuts!

Thanks again!

Find the number of selections that can be made taking 4 letters from the word "ENTRANCE".
(A) 70
(B) 36
(C) 35
(D) 72
(E) 32

Hi,

I fail to understand as to why have you considered multiple scenario's for the 4 letter selection.
I believe that the formula of nCr i.e. 8C4 would consider all of the same. Are you trying to eradicate the possibility of some duplicates in the scenario?
Could you kindly enumerate the same principle in case we have to select 3 out of 5 letters from AABCD by mentioning all the possible selections, so we can understand what is not to be considered?

Thanks

Hi,
8C4 will be correct if you have to choose 4 out of 8 different items...
But if you have few items same, then few groups become SAME and thus extra..
take this example in hand-
4 letters from the word "ENTRANCE...
In 8C4 ENTRANCE will be one and ENTRANCE will be another..
But are they different NO-- they consist of same 4 letters E, N, T, R....

There fore we require to work out individually..

select 3 out of 5 letters from AABCD..
there are four different letters - ABCD so you can choose 3 out of them in 4C3 = 4 ways..
Now let the three consist of AA, so the remaining 3rd can be any of BCD - 3 ways
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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Re: NEW SET of good PS(3) [#permalink]

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05 Jun 2016, 00:53
chetan2u

Hi,

Thanks for the explanation.

B. 2 N-s and other letters are different: 2C2(2 N-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;
C. 2 E-s and other letters are different: 2C2(2 E-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;;

In B - Can you confirm why only 5C2 is considered i.e. five distinct letters. Is it because if all 6 letters were considered, then each combination formed would have a duplicate like TE, TE or AE, AE because of 2 E's which would in turn lead to duplication.

P.S.
Its taken me over 20 mins to understand what i wanted to ask. Shoooo
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Re: NEW SET of good PS(3) [#permalink]

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05 Jun 2016, 02:54
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nishi999 wrote:
chetan2u

Hi,

Thanks for the explanation.

B. 2 N-s and other letters are different: 2C2(2 N-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;
C. 2 E-s and other letters are different: 2C2(2 E-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;;

In B - Can you confirm why only 5C2 is considered i.e. five distinct letters. Is it because if all 6 letters were considered, then each combination formed would have a duplicate like TE, TE or AE, AE because of 2 E's which would in turn lead to duplication.

P.S.
Its taken me over 20 mins to understand what i wanted to ask. Shoooo

Hi nishi999,

Why we have taken only 5C2 i.e. five distinct letters..
Hi we are looking for cases where 2 Ns are there, but if we choose 2 out of remaining 6, we will get those cases too where two Es are there..
But we are already working on BOTH 2Es and 2Ns together ...
..
so we choose first 2-Ns and other 2 different ......
then we choose where 2-Es are there and other 2 are different..
Finally we choose cases where both N and E are used twice...

If we choose 2-Ns and choose 2 from remaining 6, it will contain cases where both 2Es and 2Ns are there..
Same thing will happen when we work on 2Es, thus will result in repetitions..
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Re: NEW SET of good PS(3)   [#permalink] 05 Jun 2016, 02:54

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# NEW SET of good PS(3)

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