Math IOQM

Q1. Find the number of ways in which 30 marks can be allotted to 8 questions if each question carries at least 2 marks? S1. x1 + ... x8 = 30-16 = 14 Stars and bars, n = 14, k = 8 14+8-1C8-1 = 21C7 Q2. In an exam the maximum marks for each of three papers is 'n' and that for fourth paper is  2n. Find the number of ways in which a student can get (3n) marks. S2. Using negative binomial expansions :

 Q1.  A. How many shortest paths from X to Y? B. From AB? S1. A.  7R4U => 11!/7!4! = 330 B. 5!/3!2! * 5!/3!2! = 100 Q2. In a box, there are 10 balls: 4 red, 3 black, 2 white, 1 yellow. In how many ways can a child select 4 balls out of these 10 balls? S2. If all balls are distinct, answer is simply 10C4  = 210 If balls of same color are identical, which is typically the case, then: All 4 same: 1 3 Same, 1 different: 2*3C1 = 6 2 same, 2 same: 3C2 = 3 2 Same, 1 diff, 1 diff: 3C1*3C1 = 9 All 4 different: 1 Total: 20 Q3. There are three papers of 100 marks each. Then find the no. of ways a student can get 150 s.t. he scores at least 60% in two papers. (only integer marks are given) S3. Let' say the marks in first 2 papers are x + 60, y + 60 and in third paper z marks. So: 60 + x + 60 + y + z = 150 => x + y + z = 30 But we could have chosen any 2 papers. So there are 3C2 ways to choose the 2 papers. So we will multiply the final answer with 3. Now each of x,y,z can ...

Q1. Find the number of ways to choose an ordered pair ((a,b)) of numbers from the set (1...10) such that (|a-b| <= 5). S1. Answer: 80 Method 1: Direct counting 1: 1,2,3,4,5,6 => count:6 Similarly the count is 6 for 2,3,4,5. So total: 30*2 = 60 pairs. Then for 6,7,8,9,10 there are 5,4,3,2,1 options. So 15*2 = 30 pairs. Total: 90 But we have counted same number pairs twice. So 90-10=80 = answer Method 2: faster and better: complementary counting Total: 10x10 = 100 Subtract invalid: 1: 7,8,9,10 2: 8,9,10 3: 9,10 4: 10 Total: 10*2 = 20 100-20 = 80 = Answer Q2. Suppose that in a poll made of 150 people, the following information was obtained: 70 of them read The Hindu , 80 read The Indian Express and 50 read Deccan Herald . 30 read both The Hindu and The Indian Express ; 20 read both The Hindu and Deccan Herald and 25 read both The Indian Express and Deccan Herald . Find at most how many of them read all the three. S2. Incorrect solution first: Initially I applied PIE without thi...

Q1. Ten 1's and ten 2's are written on a blackboard. In one turn, a player may erase any two figures. If the two figures erased are identical, they are replaced with a 2. If they are different, they are replaced with a 1. The first player wins if a 1 is left at the end, and the second player wins if a 2 is left. S1. Only P2 can win. Why? 1. There are total 19 moves since in every move the count of total numbers goes down by 1. At the end we will have a single number. 2. Let's say P1 plays the first move, then he will also play the 19th move. Let C1, C2 denote count of 1s and 2s at any given time. 3. For P1 to win, after the 18th move we need C1 = 1, C2 = 1. For P2 to win C1 = 0, C2 = 2 or C1 = 2, C2 = 0. 4. Let's focus on P1's win. That means that in the initial 18 moves both C1 and C2 should decrease by 9 each. 5. What happens in every move? Case 1: Two 1s replaced by 2 => C1 -= 2, C2 += 1 Case 2: Two 2s replaced by 2 => C1 += 0, C2 -= 1 Case 3: One 1 and One...

Q1. On a board, the numbers a, b, c, d, e, f are written in a circle (say counter-clockwise). You may increase two neighbouring numbers by 1. For which case is it impossible to equalize all of them by a sequence of such steps? Select any one option a = 4, b = 10, c = 5, d = 6, e = 9, f = 2 a = 3, b = 9, c = 7, d = 6, e = 9, f = 4 a = 4, b = 11, c = 7, d = 7, e = 10, f = 2 a = 7, b = 10, c = 5, d = 8, e = 9, f = 3 S1. First a partially correct approach which works in this case but will not work always. Final sum of all numbers is even. All the increments are also even.(multiple of 2) So the original sum of all the numbers should also be even. Only option 3 fails. Hence that's the answer. But that approach fails for: a=2,b=4,c=2,d=4,e=2,f=4 So the better way to solve it is this: Let there be 2 groups:  Group 1: a,c,e Group 2: b,d,f Each time we increment, we increment by 1 in each group. So the original sum in both the groups should be same. Else it won't work. Q2. On a 10x10 boa...

Q1. While applying Menelaus on a triangle ABC, can the traversal pass through one of its vertices? S1. No. In one of the fractions numerator would become zero. In another denominator. Q2. The points (X, Y) are taken on (CA, AB) respectively of triangle ABC. If (BX) and (CY) meet at (P) and AX/CX = BY/AY = 1/2 then find BP/PX. S2. Triangle XAB, Traversal YPC. Apply Menelaus. Answer: 3/4 Q3. In △ABC the points E, F, G are on AB, BC & CA respectively, such that AE/EB = BF/FC = CG/GA = 1/3. The (K, L, M) are the intersections of (AF) & (CE); (BG) & (AF); (CE) & (BG) respectively. If ([ABC] = 1), find ([KLM] = ?) S3. Here we need to use mass points on 2 Cevians at a time. We can't apply on all 3 together since they are not concurrent. Our aim is to find [KLM] like this:  [KLM] = [ABC] - [AKC] - [ALB] - [BMC] For [AKC], use mass points on cevians AF,CE to get the ratio AK:KF. Then [CKA]:[CKF] = AK:KF But [CKA] + [CKF] = [AFC] And [AFC]/[ABC] = FC/BC So [AFC] = 3/4 So...