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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.
_________________

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...
_________________

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.

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.

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?

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.)

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
_________________

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!).

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 \(n-1C_{r-1}\).

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 \(n-1C_{r-1}\).

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.

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....

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.
_________________

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.
_________________

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Term 1 has begun. If you're confused, wondering what my post on the last 2 official weeks was, that was pre-term. What that means is that the school...