Showing posts with label Combinatorics. Show all posts
Showing posts with label Combinatorics. Show all posts
Wednesday, October 12, 2011
Problem of the Day 12 Oct 2011
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Monday, October 10, 2011
Problem of the Day 10 Oct 2011
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Saturday, October 8, 2011
Problem of the day 8th oct 2011
Rohan is asked to figure out the marks scored by Sunil in three different subjects with the help of
certain clues. He is told that the product of the marks obtained by Sunil is 72 and the sum of the
marks obtained by Sunil is equal to the Rohan’s current age (in completed years). Rohan could not
answer the question with this information. When he was also told that Sunil got the highest marks
in Physics among the three subjects, he immediately answered the question correctly. What is the
sum of the marks scored by Sunil in the two subjects other than Physics?
(a) 6 (b) 8 (c) 10 (d) Cannot be determined
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Friday, October 7, 2011
Problem of the day 7th Oct 2011
Ten books are arranged in a row on a bookshelf. A student has to select three out of these ten books
in such a way that no two books selected by him must have been lying adjacently. In how many
ways can he make the selection?
(a) 56 (b) 64 (c) 72 (d) None of these
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Tuesday, September 27, 2011
Permutations and Combinations- Free Lecture Notes
Many of you have emailed about some easy reference for Permutations and Combinations, Number theory etc. I cannot say if it is an easy reference, but it is one concise one.. It is from one of our Professors at IIT Kanpur, Dr A K Lal, great chap, I must say. This is one gem of a compilation. Hope it helps you.. If the material is too tough, just ignore!
http://home.iitk.ac.in/~arlal/book/mth202.pdf
Enjoy!
http://home.iitk.ac.in/~arlal/book/mth202.pdf
Enjoy!
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Sunday, September 25, 2011
Problem of the Day 25 Sept 2011
Let P be the set of all the vertices of a regular polygon of 25 sides with its center at C. How many triangles have vertices in P and contain the point C in the interior of the triangles?
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Thursday, September 15, 2011
Problem of the day 15 sept 2011
Rachel and Brian are playing a game in a grid with 1 row of 2011 squares. Initially, there is one
white checker in each of the first two squares from the left, and one black checker in the third square
from the left. At each stage, Rachel can choose to either run or ght. If Rachel runs, she moves the
black checker 1 unit to the right, and Brian moves each of the white checkers one unit to the right. If
Rachel chooses to fight, she pushes the checker immediately to the left of the black checker 1 unit to
the left, the black checker is moved 1 unit to the right, and Brian places a new white checker in the
cell immediately to the left of the black one. The game ends when the black checker reaches the last
cell. How many different final configurations are possible?
a) 2011 b) 2010 c) 2009 d) None
white checker in each of the first two squares from the left, and one black checker in the third square
from the left. At each stage, Rachel can choose to either run or ght. If Rachel runs, she moves the
black checker 1 unit to the right, and Brian moves each of the white checkers one unit to the right. If
Rachel chooses to fight, she pushes the checker immediately to the left of the black checker 1 unit to
the left, the black checker is moved 1 unit to the right, and Brian places a new white checker in the
cell immediately to the left of the black one. The game ends when the black checker reaches the last
cell. How many different final configurations are possible?
a) 2011 b) 2010 c) 2009 d) None
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Saturday, July 25, 2009
Problem of the day 25.07.09
In 1896 lord Coin has decided to play a game. From the January 1 till December 31 every day he chooses among two match boxes an arbitrary one and placed a match from it to another box (if the chosen box was not empty). If the chosen box was empty then he placed a match from
the other box to the chosen one. What is the probability that after the December 31 the both boxes will have an equal number of matches if at the beginning each box had a) n = 400 b) n = 200 c) n = 100 matches?
the other box to the chosen one. What is the probability that after the December 31 the both boxes will have an equal number of matches if at the beginning each box had a) n = 400 b) n = 200 c) n = 100 matches?
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Monday, July 13, 2009
Problem of The day 13.07.09
On a circle 26 equidistant points are marked. these points are joined to form a triangles. Of the triangles formed, how many of them will have their circumcenter on one of their sides.?
A) 318 B) 312 C) 288 d) 624 e) None of these
A) 318 B) 312 C) 288 d) 624 e) None of these
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Wednesday, July 8, 2009
Problems 8.07.09
I could not post due to some engagements. Here are a bunch of problems to compensate :)
Question 1)
A + B + C + D = D + E + F + G = G + H + I = 17 where each letter represent a number from 1 to 9. Find out number of ordered pairs (D,G) if letter A = 4.
a) 0 b) 1 c)2 d) 3 e) none of these
Question 2)
The sequence 1, 3, 4, 9, 10, 12..... includes all numbers that are a sum of one or more distinct powers of 3. Then the 50th term of the sequence is
a. 252 b. 283 c. 327 d. 360 e) none of these
Question 3)
Given that g(h(x)) = 2x² + 3x and h(g(x)) = x² + 4x − 4 for all
real x. WHich of the following could be the value of g(-4)?
a)1 b) -1 c) 2 d) -2 e) -3
Question 4)
If a, x, b and y are real numbers and ax+by = 4 and ax² +by² = 2 and
ax³ + by³= −3 then find (2x − 1)(2y − 1)
a)4 b) 3 c) 5 d) -3 e) cannot be determined.
Question 5)
K1,K2,K3...K30 are thirty toffees. A child places these toffees on a circle, such that there are exactly n ( n is a positive integer) toffees placed between Ki and Ki+1 and no two toffees overlap each other. Find n
a)4 b) 5 c) 9 d) 12 e) 13
Question 6)
For the n found in previous question, which of the two toffees are adjacently
placed on the circle? ( All other conditions remaining same)
a) K11 and K13 b) K6 and K23 c) K2 and K10 d) K11 and K18
e) K20 and K28
Question 1)
A + B + C + D = D + E + F + G = G + H + I = 17 where each letter represent a number from 1 to 9. Find out number of ordered pairs (D,G) if letter A = 4.
a) 0 b) 1 c)2 d) 3 e) none of these
Question 2)
The sequence 1, 3, 4, 9, 10, 12..... includes all numbers that are a sum of one or more distinct powers of 3. Then the 50th term of the sequence is
a. 252 b. 283 c. 327 d. 360 e) none of these
Question 3)
Given that g(h(x)) = 2x² + 3x and h(g(x)) = x² + 4x − 4 for all
real x. WHich of the following could be the value of g(-4)?
a)1 b) -1 c) 2 d) -2 e) -3
Question 4)
If a, x, b and y are real numbers and ax+by = 4 and ax² +by² = 2 and
ax³ + by³= −3 then find (2x − 1)(2y − 1)
a)4 b) 3 c) 5 d) -3 e) cannot be determined.
Question 5)
K1,K2,K3...K30 are thirty toffees. A child places these toffees on a circle, such that there are exactly n ( n is a positive integer) toffees placed between Ki and Ki+1 and no two toffees overlap each other. Find n
a)4 b) 5 c) 9 d) 12 e) 13
Question 6)
For the n found in previous question, which of the two toffees are adjacently
placed on the circle? ( All other conditions remaining same)
a) K11 and K13 b) K6 and K23 c) K2 and K10 d) K11 and K18
e) K20 and K28
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Thursday, October 16, 2008
Mini Concept: Multinomial Theorem
Some Trickery of Multinomial Theorem
x+y+z=n the number of non negative solutions of this equation will be
(n+3-1)C(3-1)=(n+2)C2
if we extend this to r variables then the formula becomes (n+r-1)C(r-1)
if we remove 0, means we need only positive integral solutions to the equation then we get formula as (n-1)C(r-1)
Lets take up one example.
On the occasion of Diwali, PAPA CHIPS is offering one of five prizes with every packet( the prize is inside the packet). the prizes include a pen, pencil, a CD, a movie ticket and a small game. Banta Singh is a fan of PAPA chips and he keeps buying the chips, what is the probability that Banta Singh gets all the five prizes by buying 12 packets of PAPA chips
Solution:
Chuck the story, the question is there are 5 variables and we need the solutions to the equation
a+b+c+d+e=12 ( non negative)
and a+b+c+d+e=12 ( positive integral)
The first case comes as every packet has a prize, and those 5 are the only kinds of prizes.
Second comes from that we need each kind of prize.
so the answer is 11C4/16C4=11!12!/(7!16!)=11.10.9.8/13.14.15.16= 33/182
Lets take another example
If the sum of 101 distinct terms in arithmetic progression is zero , in how many ways can three of these terms be selected such that their sum is zero?
Solution
it is obvious that the middle term is zero
so a(51)=0
so the terms are
-50D, -49D,....,-D, 0, D, ....49D, 50D
now the sum of 3 numbers to be zero
Case 1) if we pick 0, then we have to pick one positive and one negative, which must be equal except for the sign . so 50 ways
case 2) we leave 0 and pick two positive and one negative
then xD+yD-zD=0
x+y=z
z can vary from 1 to 50
we need positive solutions to the equation
which comes 0C1+1C1+2C1...+49C1
add this it will come to 50C2
case 3 it will be same as case 2
we get 50C2
hence total is 2.50C2+50=2500
x+y+z=n the number of non negative solutions of this equation will be
(n+3-1)C(3-1)=(n+2)C2
if we extend this to r variables then the formula becomes (n+r-1)C(r-1)
if we remove 0, means we need only positive integral solutions to the equation then we get formula as (n-1)C(r-1)
Lets take up one example.
On the occasion of Diwali, PAPA CHIPS is offering one of five prizes with every packet( the prize is inside the packet). the prizes include a pen, pencil, a CD, a movie ticket and a small game. Banta Singh is a fan of PAPA chips and he keeps buying the chips, what is the probability that Banta Singh gets all the five prizes by buying 12 packets of PAPA chips
Solution:
Chuck the story, the question is there are 5 variables and we need the solutions to the equation
a+b+c+d+e=12 ( non negative)
and a+b+c+d+e=12 ( positive integral)
The first case comes as every packet has a prize, and those 5 are the only kinds of prizes.
Second comes from that we need each kind of prize.
so the answer is 11C4/16C4=11!12!/(7!16!)=11.10.9.8/13.14.15.16= 33/182
Lets take another example
If the sum of 101 distinct terms in arithmetic progression is zero , in how many ways can three of these terms be selected such that their sum is zero?
Solution
it is obvious that the middle term is zero
so a(51)=0
so the terms are
-50D, -49D,....,-D, 0, D, ....49D, 50D
now the sum of 3 numbers to be zero
Case 1) if we pick 0, then we have to pick one positive and one negative, which must be equal except for the sign . so 50 ways
case 2) we leave 0 and pick two positive and one negative
then xD+yD-zD=0
x+y=z
z can vary from 1 to 50
we need positive solutions to the equation
which comes 0C1+1C1+2C1...+49C1
add this it will come to 50C2
case 3 it will be same as case 2
we get 50C2
hence total is 2.50C2+50=2500
Wednesday, October 15, 2008
Problem Of The Week 53
How many ways can a size k + 1 subset with maximum element m + 1 can be created from the given set S={1,2,3,.....,n+1} ?
Wednesday, October 8, 2008
Problem Of The Week 48
If square tiles are to be fitted on a square floor in such a way that the floor looks the same from all the sides. The tiles are availble in 6 different colors. In how many ways can the floor be made if it should have maximum possible colors?
1) 360
2) 60
3) 1296
4) 256
5) None of these
1) 360
2) 60
3) 1296
4) 256
5) None of these
Thursday, October 2, 2008
Problem Of The Week 37
What is the sum of the smallest and the largest number of fridays the 13th that can occur in any year ?
Sunday, September 28, 2008
Problem Of The Week 35
How many 3-d igit numbers are such that one of the digits is the average of the other two?
(A) 96 (B) 112 (C) 120 (D) 104 (E) 256
(A) 96 (B) 112 (C) 120 (D) 104 (E) 256
Problem Of The Week 33
The sum of base-10 logarithms of divisors of 10^n is 792. what is n?
(A) 11 (B) 12 (C) 10 (D) 13 (E) 14
Wednesday, September 24, 2008
Problem Of The Week 25
Let X and Y be distinct 3 digit palindromes such that X>Y. How many pairs (X,Y) exist such that X-Y is also a 3 digit palindrome ?
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