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Re: NEW SET of good PS(3)
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18 Mar 2013, 13:24
annutalreja wrote: Hi, still not clear why answer to the 2nd ques should be P1 and not P2. Shouldn't both P and 1 be deducted from the set? p yes, but not 1. Consider this, say p=7 how many numbers are less than 7 having no common factors with 7 other than 1: 1, 2, 3, 4, 5, 6 > 71=6. (7 and 1 do not share any common factor other than 1.) Completes solution is here: newsetofgoodps8544060.html#p1102100Hope it helps.
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30 Nov 2013, 22:17
VeritasPrepKarishma wrote: yogesh1984 wrote: Bunuel, 1 Collinear point issue will arise in case of overlapping values of x, y ? (as in here we have all the overlapping range for x & y). Also since range here is small for both x, y (ie.=3) we can manually calculate the collinear points but in case of large range how do we go about it ? should it be = # overlapping points on X + # overlapping points on Y + # diagonal points (which will essentially be min(# overlapping points on X , Y) 1 ) Not so sure on this though ... 2 I see a similar Question in OG12 PS Q.229 The method explained here in the above example does not seems to fit too well there. basically in the question we have 4 <= X <=5, 6<= Y <=16. Can you please throw some light in the context of OG question.... TIA ~ Yogesh Check out this thread: psrighttrianglepqr71597.html?hilit=how%20many%20triangles#p830694It discusses what to do in case of a larger range. Hi Karishma, To find out the possible number of right triangles I tried as below: No. of rectangles *4 (for each orientation) For example: How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 2 ≤ x ≤ 4 and 5 ≤ y ≤ 7? No. of rectangles = 9 (this by actual counting of rectangles) No. of right triangles = 4*9 = 36, Is this correct? Secondly, I tried to calculate no. of rectangles via combinations but i count understand why no. of rectangles would be 3c2*3c2? To me it should be = 3c2*(31)=6, but this is not correct.



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Re: NEW SET of good PS(3)
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01 Dec 2013, 00:28
yossarian84 wrote: 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
Answer: C.
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 \(x8\) vouchers, so that each can get from zero to \(x8\) as at "least 2", or 2*4=8, we already booked. Let \(x8\) 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,
\(ttttt\) 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 \((5+41)C(41)=8C3\).
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 \((K+41)C(41)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\).
\((k+1)(k+2)(k+3)=3!*120=720\). > \(k=7\). Plus the 8 tickets we booked earlier: \(x=k+8=7+8=15\).
Answer: C (15).
P.S. 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 \(n+r1C_{r1}\).
The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n1C_{r1}\).
Hope it helps. Awesome...hats off...this is totally new to me...widens my realm..and strengthens my reasoning...thanks a lot Karishma/Bunnel, There one more formula to distribute n things in m, such that each m can receive 0 to n items = m^n. Thing to be distributed goes in the power. Can we do 4^(n8)=120??



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Re: NEW SET of good PS(3)
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01 Dec 2013, 20:35
cumulonimbus wrote: VeritasPrepKarishma wrote: yogesh1984 wrote: Bunuel, 1 Collinear point issue will arise in case of overlapping values of x, y ? (as in here we have all the overlapping range for x & y). Also since range here is small for both x, y (ie.=3) we can manually calculate the collinear points but in case of large range how do we go about it ? should it be = # overlapping points on X + # overlapping points on Y + # diagonal points (which will essentially be min(# overlapping points on X , Y) 1 ) Not so sure on this though ... 2 I see a similar Question in OG12 PS Q.229 The method explained here in the above example does not seems to fit too well there. basically in the question we have 4 <= X <=5, 6<= Y <=16. Can you please throw some light in the context of OG question.... TIA ~ Yogesh Check out this thread: psrighttrianglepqr71597.html?hilit=how%20many%20triangles#p830694It discusses what to do in case of a larger range. Hi Karishma, To find out the possible number of right triangles I tried as below: No. of rectangles *4 (for each orientation) For example: How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 2 ≤ x ≤ 4 and 5 ≤ y ≤ 7? No. of rectangles = 9 (this by actual counting of rectangles) No. of right triangles = 4*9 = 36, Is this correct? Secondly, I tried to calculate no. of rectangles via combinations but i count understand why no. of rectangles would be 3c2*3c2? To me it should be = 3c2*(31)=6, but this is not correct. The method is correct for this question. Number of right triangles will be 36.
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Re: NEW SET of good PS(3)
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01 Dec 2013, 20:45
cumulonimbus wrote: VeritasPrepKarishma wrote: yogesh1984 wrote: Bunuel, 1 Collinear point issue will arise in case of overlapping values of x, y ? (as in here we have all the overlapping range for x & y). Also since range here is small for both x, y (ie.=3) we can manually calculate the collinear points but in case of large range how do we go about it ? should it be = # overlapping points on X + # overlapping points on Y + # diagonal points (which will essentially be min(# overlapping points on X , Y) 1 ) Not so sure on this though ... 2 I see a similar Question in OG12 PS Q.229 The method explained here in the above example does not seems to fit too well there. basically in the question we have 4 <= X <=5, 6<= Y <=16. Can you please throw some light in the context of OG question.... TIA ~ Yogesh Check out this thread: psrighttrianglepqr71597.html?hilit=how%20many%20triangles#p830694It discusses what to do in case of a larger range. Secondly, I tried to calculate no. of rectangles via combinations but i count understand why no. of rectangles would be 3c2*3c2? To me it should be = 3c2*(31)=6, but this is not correct. Make the coordinate axis, mark the coordinates and make all vertical and horizontal lines, To make a rectangle, you need 2 horizontal and 2 vertical lines. You have 3 vertical lines and 3 horizontal lines. You select 2 of each in 3C2 * 3C2 ways. Similarly, say 2 <= x <= 5 and 5<= y <= 7. Calculate manually to get 18 rectangles. Use 4C2*3C2 to get 18 rectangles too.
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Re: NEW SET of good PS(3)
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01 Dec 2013, 20:50
cumulonimbus wrote: Karishma/Bunnel,
There one more formula to distribute n things in m, such that each m can receive 0 to n items = m^n. Thing to be distributed goes in the power.
Can we do 4^(n8)=120?? Things need to be distinct in that case. This is a case of identical things. Check these posts for a discussion on these differences: http://www.veritasprep.com/blog/2011/12 ... 93part1/http://www.veritasprep.com/blog/2011/12 ... spartii/
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Re: NEW SET of good PS(3)
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05 Dec 2013, 20:11
The method is correct for this question. Number of right triangles will be 36.
Hi Karishma, Do you mean I cannot use the above method for all questions for getting the number of right triangles?



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Re: NEW SET of good PS(3)
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05 Dec 2013, 20:33
cumulonimbus wrote: The method is correct for this question. Number of right triangles will be 36.
Hi Karishma, Do you mean I cannot use the above method for all questions for getting the number of right triangles? Whenever an innovative method is used, we need to understand its assumptions. Here we are looking for right triangles with all integer coordinates lying within a certain range. If one of these conditions is not met, the method will change. Of course there is nothing special about these particular values: 2 ≤ x ≤ 4 and 5 ≤ y ≤ 7 and as discussed above, the method will work for any other such set of values e.g. 2 <= x <= 5 and 5<= y <= 7 or 1 <= x <= 7 and 2<= y <= 7 etc. There is nothing wrong with this approach  in fact it's great. You can certainly solve a similar question using this approach. But it is important for you to understand the generic method too (for more generic questions, say all triangles, not just right triangles or some other variation) so that you can twist it according to the question at hand, if required.
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Re: NEW SET of good PS(3)
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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) P1 (B) P2 (C) (P+1)/2 (D) (P1)/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 n1(as n is not included) and another 1 (as 1 is a common factor, Thus n2.
Apologies if I am wrong. PLease correct me .



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Re: NEW SET of good PS(3)
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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) P1 (B) P2 (C) (P+1)/2 (D) (P1)/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 n1(as n is not included) and another 1 (as 1 is a common factor, Thus n2.
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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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: NEW SET of good PS(3)
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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 p1 only.[/quote]
Can anyone please explain from where this (X+3) came? And how (x+3)C3=120 is equal to x+3 = 10?
Thanks in advance.



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Re: NEW SET of good PS(3)
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10 Feb 2016, 02:00
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 p1 only. Can anyone please explain from where this (X+3) came? And how (x+3)C3=120 is equal to x+3 = 10?Thanks in advance.[/quote] Hi, for your Q: 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+33)!= (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 Ns and other letters are different; C. 2 Es and other letters are different; D. 2 Ns, 2 Es. 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 Ns and other letters are different: 2C2(2 Ns out of 2)*5C2(other 2 letters from distinct 5 letters left)=10; C. 2 Es and other letters are different: 2C2(2 Es out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;; D. 2 Ns, 2 Es: 2C2*2C2=1. 15+10+10+1=36 Answer: B. 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)
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05 Jun 2016, 00:03
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 ENTR ANCE will be one and ENTR ANCE 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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Re: NEW SET of good PS(3)
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05 Jun 2016, 00:53
chetan2uHi, Thanks for the explanation. B. 2 Ns and other letters are different: 2C2(2 Ns out of 2)*5C2(other 2 letters from distinct 5 letters left)=10; C. 2 Es and other letters are different: 2C2(2 Es 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)
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05 Jun 2016, 02:54
nishi999 wrote: chetan2uHi, Thanks for the explanation. B. 2 Ns and other letters are different: 2C2(2 Ns out of 2)*5C2(other 2 letters from distinct 5 letters left)=10; C. 2 Es and other letters are different: 2C2(2 Es 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 2Ns and other 2 different ...... then we choose where 2Es are there and other 2 are different.. Finally we choose cases where both N and E are used twice... If we choose 2Ns 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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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 OPEN DISCUSSION OF HIS QUESTION IS HERE: https://gmatclub.com/forum/abcdeisar ... 86284.html2. 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) P1 (B) P2 (C) (P+1)/2 (D) (P1)/2 (E) 2 OPEN DISCUSSION OF HIS QUESTION IS HERE: https://gmatclub.com/forum/thefunction ... 03852.html3. How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72 OPEN DISCUSSION OF HIS QUESTION IS HERE: https://gmatclub.com/forum/howmanynum ... 26647.html4.A certain quantity is measured on two different scales, the Rscale and the Sscale, that are related linearly. Measurements on the Rscale of 6 and 24 correspond to measurements on the Sscale of 30 and 60, respectively. What measurement on the Rscale corresponds to a measurement of 100 on the Sscale? (A) 20 (B) 36 (C) 48 (D) 60 (E) 84 OPEN DISCUSSION OF HIS QUESTION IS HERE: https://gmatclub.com/forum/acertainqu ... 96389.html5. 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 OPEN DISCUSSION OF HIS QUESTION IS HERE: https://gmatclub.com/forum/mrssmithha ... 98225.html6. 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) OPEN DISCUSSION OF HIS QUESTION IS HERE: https://gmatclub.com/forum/thisyearhe ... 00891.html7. 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 OPEN DISCUSSION OF HIS QUESTION IS HERE: https://gmatclub.com/forum/beforebeing ... 25936.html8. 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 OPEN DISCUSSION OF HIS QUESTION IS HERE: https://gmatclub.com/forum/howmanypos ... 02927.html9. 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. OPEN DISCUSSION OF HIS QUESTION IS HERE: https://gmatclub.com/forum/findthenum ... 98237.html10. 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 OPEN DISCUSSION OF HIS QUESTION IS HERE: https://gmatclub.com/forum/howmanytri ... 98236.html
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Re: NEW SET of good PS(3) &nbs
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07 Apr 2018, 14:09



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