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# NEW SET of good PS(3)

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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
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Re: NEW SET of good PS(3) [#permalink]

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22 Jan 2012, 05:08
akhileshankala wrote:
It would like to point out tht the resoning given is wrong. the triplets need not necessarily be adjacent. tht's the flaw.
my way:
no: of collinear points=?
horizontal and vertical lines both give the same no: and each line of 6 points gives 6C3 possibs.
hence horz and vert. lines give a total of 2*6*6C3.
next 2 diagonals give same no: of such possibs.
consider any diagonal direction. it gives 3,4,5,6,5,4,3 collinear points along 6 parallel lines corresponding to any diagonalic direction and each of the points gives us their corresponding triples-3C3+4C3+5C3+6C3+5C3+4C3+3C3.

along 2 such dirs. this adds up to 2*(2*(3C3+4C3+5C3)+6C3).

total no: of line forming selections= 2*6*6C3+ 2*(2*(3C3+4C3+5C3)+6C3).

Yes, I did miss out on the non-adjacent collinear points! And on the face of it, your calculation looks correct. I will put some more time on this variation tomorrow (since today is Sunday!) and get back if needed.
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Karishma
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 12 Dec 2010 Posts: 279 Concentration: Strategy, General Management GMAT 1: 680 Q49 V34 GMAT 2: 730 Q49 V41 GPA: 4 WE: Consulting (Other) Followers: 9 Kudos [?]: 50 [0], given: 23 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 14 Mar 2012, 06:37 akhileshankala wrote: VeritasPrepKarishma wrote: dimri10 wrote: if anyone can help please to clarify the methos: let's say that the Q was: 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≤6 and 1≤y≤6? how will it be solved: will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different. thank's in advance I think your question is quite similar to yogesh1984's question above. I missed answering his question (thought of doing it later due to the diagram involved but it skipped my mind). Anyway, let me show you how I would solve such a question. Both the questions can be easily answered using this method. 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≤6 and 1≤y≤6? Ok, so we have a total of 36 co-ordinates (as shown below by the red and black dots). We need to make triangles so we need to select a triplet of co-ordinates out of these 36 which can be done in 36C3 ways. Out of these, we need to get rid of those triplets where the points are collinear. How many such triplets are there? Look at the diagram: Attachment: Ques2.jpg The Black dots are the outermost points. Red dots are the inside points. Now each of these red dots is the center point for 4 sets of collinear points (as shown by the red arrows). Hence the 4*4 = 16 red dots will make 16*4 = 64 triplets of collinear points. These 64 triplets account for all collinear triplets except those lying on the edges. Each of the 4 edges will account for 4 triplets of collinear points shown by the black arrows. Hence, there will be another 4*4 = 16 triplets of collinear points. Total triplets of collinear points = 64 + 16 = 80 Therefore, total number of triangles you can make = 36C3 - 80 Similarly you can work with 1<=x<=5 and -9<=y<=3. The number of red dots in this case = 11*3 = 33 So number of collinear triplets represented by red arrows will be = 33*4 = 132 Number of black arrows will be 3 + 11 + 3 + 11 = 28 Total triplets of collinear points = 132 + 28 = 160 Total triangles in this case = 65C3 - 160 Ma'am, It would like to point out tht the resoning given is wrong. the triplets need not necessarily be adjacent. tht's the flaw. my way: no: of collinear points=? horizontal and vertical lines both give the same no: and each line of 6 points gives 6C3 possibs. hence horz and vert. lines give a total of 2*6*6C3. next 2 diagonals give same no: of such possibs. consider any diagonal direction. it gives 3,4,5,6,5,4,3 collinear points along 6 parallel lines corresponding to any diagonalic direction and each of the points gives us their corresponding triples-3C3+4C3+5C3+6C3+5C3+4C3+3C3. along 2 such dirs. this adds up to 2*(2*(3C3+4C3+5C3)+6C3). total no: of line forming selections= 2*6*6C3+ 2*(2*(3C3+4C3+5C3)+6C3). Can you please elaborate on the bolded part in details... _________________ My GMAT Journey 540->680->730! ~ When the going gets tough, the Tough gets going! Manager Joined: 09 Mar 2012 Posts: 97 Location: India GMAT 1: 740 Q50 V39 GPA: 3.4 WE: Business Development (Manufacturing) Followers: 2 Kudos [?]: 18 [0], given: 12 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 14 Mar 2012, 08:48 This is a 6x6 square. For each diagonal of this square, you have 8 parallel lines, you can draw within the square by joining the vertices that lies on the edges of the square. eg: Join (1,2) & (2,1); (1,3) & (3,1); (1,4) & (4,1); (1,5) & (5,1); to get 4 parallel lines along the diagonal (1,6)-(6,1) Similarly you can get 4 lines on the other side of the diagonal. Of these, (line joining (1,2) to (2,1) is of no use to us since it contains only 2 points within the square) the line joining point (1,3) & (3,1) contains total of 3 integer co-ordinates, the line joining point (1,4) & (4,1) contains total of 4 integer co-ordinates, and so on..... Any 3 points that you select from these lines will be collinear and not form a traingle. Thus, you have 3,4,5,6,5,4,3 points collinear along the lines parallel to the diagonal. Rest as akhilesh has mentioned. You may draw a figure by plotting these points. My 1st post on this forum, so Apologies for the weird explanation. Senior Manager Joined: 12 Dec 2010 Posts: 279 Concentration: Strategy, General Management GMAT 1: 680 Q49 V34 GMAT 2: 730 Q49 V41 GPA: 4 WE: Consulting (Other) Followers: 9 Kudos [?]: 50 [0], given: 23 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 19 Mar 2012, 09:56 VeritasPrepKarishma wrote: yogesh1984 wrote: Can you please elaborate on the bolded part in details... Check out this post. I have explained this question in detail in this post. It fixes the problem my above given solution had. http://www.veritasprep.com/blog/2011/09 ... o-succeed/ Aah that one is awesome !! bole to crystal clear now _________________ My GMAT Journey 540->680->730! ~ When the going gets tough, the Tough gets going! Intern Joined: 01 Aug 2011 Posts: 23 Followers: 0 Kudos [?]: 7 [0], given: 15 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 06 Jul 2012, 02:12 Shouldn't the answer to Question 2 be B? Math Expert Joined: 02 Sep 2009 Posts: 39037 Followers: 7750 Kudos [?]: 106470 [0], given: 11626 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 06 Jul 2012, 02:21 shreya717 wrote: Shouldn't the answer to Question 2 be B? Answers are given in the following post: new-set-of-good-ps-85440.html#p642321 OA for this question is A, not B. 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 a prime number then f(p)= A. p-1 B. p-2 C. (p+1)/2 D. (p-1)/2 E. 2 If not the wording the question wouldn't be as tough as it is now. The GMAT often hides some simple concept in complicated way of delivering it. This question for instance basically asks: how many positive integers are less than given prime number p which have no common factor with p except 1. Well as p is a prime, all positive numbers less than p have no common factors with p (except common factor 1). So there would be p-1 such numbers (as we are looking number of integers less than p). For example: if p=7 how many numbers are less than 7 having no common factors with 7: 1, 2, 3, 4, 5, 6 --> 7-1=6. Answer: A. Hope it's clear. _________________ Intern Joined: 20 Jun 2011 Posts: 46 Followers: 1 Kudos [?]: 102 [0], given: 1 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 16 Oct 2012, 07:07 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 $$x-8$$ 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. 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. I tried to backsolve, but even after I saw the formula I failed miserably. Is it possible to be solving this through backsolving, or would that be too cumbersome? If it is possible, how would you then proceed? Thanks! Manager Joined: 20 Feb 2013 Posts: 141 GMAT 3: 720 Q48 V42 Followers: 9 Kudos [?]: 36 [0], given: 24 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 18 Mar 2013, 14:15 Hi, still not clear why answer to the 2nd ques should be P-1 and not P-2. Shouldn't both P and 1 be deducted from the set? Math Expert Joined: 02 Sep 2009 Posts: 39037 Followers: 7750 Kudos [?]: 106470 [0], given: 11626 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 18 Mar 2013, 14:24 annutalreja wrote: Hi, still not clear why answer to the 2nd ques should be P-1 and not P-2. 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 --> 7-1=6. (7 and 1 do not share any common factor other than 1.) Completes solution is here: new-set-of-good-ps-85440-60.html#p1102100 Hope it helps. _________________ Manager Joined: 07 Feb 2011 Posts: 103 Followers: 0 Kudos [?]: 60 [0], given: 45 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 26 Mar 2013, 11:38 Bunuel are these all official? The first problem uses the word different which I think is, different from, the word 'unique'. Subtle difference I never noticed but huge implications _________________ We appreciate your kudos' Current Student Joined: 06 Jan 2013 Posts: 29 Schools: CBS '18 (M) GMAT 1: 700 Q43 V42 GMAT 2: 730 Q47 V42 Followers: 1 Kudos [?]: 15 [0], given: 70 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 06 Apr 2013, 10:04 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? Thanks in advance That's a notation for Combinations. You have six points to choose from and you're choosing three so (6!)/(3!*3!). Hope that's helpful. Intern Joined: 25 May 2013 Posts: 28 Followers: 0 Kudos [?]: 26 [0], given: 16 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 04 Jul 2013, 08:40 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 $$x-8$$ 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. 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. Can you explain this more? Math Expert Joined: 02 Sep 2009 Posts: 39037 Followers: 7750 Kudos [?]: 106470 [0], given: 11626 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 04 Jul 2013, 08:48 akijuneja 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 $$x-8$$ 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. 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. Can you explain this more? Check here: mrs-smith-has-been-given-film-vouchers-each-voucher-allows-98225.html Similar questions: larry-michael-and-doug-have-five-donuts-to-share-if-any-108739.html how-many-positive-integers-less-than-10-000-are-there-in-85291.html how-many-positive-integers-less-than-10-000-are-there-in-whi-91261.html Hope it helps. _________________ Manager Joined: 07 May 2013 Posts: 109 Followers: 0 Kudos [?]: 28 [0], given: 1 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 22 Nov 2013, 22:00 Buneul, please explain the solution for problem no.9. Thanks in advance . Math Expert Joined: 02 Sep 2009 Posts: 39037 Followers: 7750 Kudos [?]: 106470 [0], given: 11626 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 23 Nov 2013, 06:10 madn800 wrote: Buneul, please explain the solution for problem no.9. Thanks in advance . Please read the whole thread before posting a question: new-set-of-good-ps-85440-20.html#p652346 _________________ Manager Joined: 14 Nov 2011 Posts: 150 Location: United States Concentration: General Management, Entrepreneurship GPA: 3.61 WE: Consulting (Manufacturing) Followers: 0 Kudos [?]: 18 [0], given: 103 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 01 Dec 2013, 01:28 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 $$x-8$$ 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. 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. 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^(n-8)=120?? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7380 Location: Pune, India Followers: 2291 Kudos [?]: 15146 [0], given: 224 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 01 Dec 2013, 21:35 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: ps-right-triangle-pqr-71597.html?hilit=how%20many%20triangles#p830694 It 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*(3-1)=6, but this is not correct. The method is correct for this question. Number of right triangles will be 36. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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01 Dec 2013, 21:50
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^(n-8)=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 ... 93-part-1/
http://www.veritasprep.com/blog/2011/12 ... s-part-ii/
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Veritas Prep | GMAT Instructor
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 14 Nov 2011 Posts: 150 Location: United States Concentration: General Management, Entrepreneurship GPA: 3.61 WE: Consulting (Manufacturing) Followers: 0 Kudos [?]: 18 [0], given: 103 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 05 Dec 2013, 21:11 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? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7380 Location: Pune, India Followers: 2291 Kudos [?]: 15146 [0], given: 224 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 05 Dec 2013, 21:33 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: NEW SET of good PS(3)   [#permalink] 05 Dec 2013, 21:33

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