Math IOQM

Q. The six faces of a cube are coloured using all three colours red, blue and green. Two colourings are considered the same if one can be obtained from the other by rotating the cube. How many distinct colourings are there? S21. Since all 3 colors need to be used, there are 3 cases: 4,1,1 3,2,1 2,2,2 Case 1: 4,1,1 3 ways to choose the color with 4 faces. Now the remaining 2 faces with 2 different colors have 2 options, opposite or adjacent. So 3*2 = 6 ways. Red Blue Yellow Yellow Yellow Yellow Drag the cube. The red and blue faces are adjacent. The other four faces have the same colour. ↶ Left ↑ Up Reset ↓ Down Right ↷ Case 2: 3,2,1 Choose first color in 3 ways, second in 2 and third in 1 way: 3*2*1 = 6 total. For each of them 3 sub cases as mentioned below...

Q13. How many sets of three edges of a cube can be chosen so that no two chosen edges share a vertex? S13. 12 ways to choose the first edge. Each of the 2 vertices of this edge had 2 more edges to it. So they are excluded from future selections. 7 edges and 6 vertices remain. Out of the 6 vertices, 2 have 3 edges each and 4 have 2 edges each. Total ways to select 2 edges from 7 = 7C2 = 21. From this we have to subtract those edge pairs which share a vertex. Consider a vertex with degree 3. It gives us 3C2 such pairs. And there are 2 such vertices. Total 6 such edge pairs. Similarly vertices with degree 2 give us 2C2*4 = 4 such pairs. 21 - 6 - 4 = 11 valid edge pairs. So total: 12*11 = 132 valid pairs. But in the first step we could have any of the 3 edges as the first edge. Later we just counted pairs. So 132/3 = 44 = answer. Now let's modify the original problem. What if we need to count any pair of edges from a cube which don't share a vertex? Each vertex has degree 3. So 8*3...

Q1. How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)  S1. 7!/3!2! Q2. Seven 6-sided dice are rolled. The probability that the sum of the numbers on the top faces is 10 can be written as n/6^7 where n is a positive integer. What is n? S2. x1+x2..x7 = 10 Each of them is minimum 1. For stars and bars: x1+x2...x7 = 3 9C6 ways out of 6^7 ways. 84/6^7 n = 84 Q3. Suzanne went to the bank and withdrew 800 dollars. The teller gave her this amount using 20 dollars bills, 50 dollars bills, and 100 dollars bills, with at least one of each denomination. How many different collections of bills could Suzanne have received? S3. 20x + 50y + 100z = 800 2x + 5y + 10z = 80 5y will be even, y = 2k 2x + 10k + 10z = 80 x + 5k + 5z = 40 x = 5(8 - k - z) x >= 1 => 8-k-z > 0 => k+z <8 => k+z <= 7 For k+z <= 7 with k>=1 and z>=1 how many ...

1. How many ways to distribute 10 different balls into 3 different bags such that none of the bags is empty? S1. Distinct objects/distinct buckets => PIE(Principle of Inclusion Exclusion) Step 1: Total cases: 3^10 since each ball can go into any of the bags. Step 2: Subtract those cases where one bag is empty: 3C1 to choose one bag. And now each ball has 2 options: 2^10. So total 3*2^10 cases need to be subtracted to account for one empty bucket. In step 2 consider what happens in detail: When we made one bucket empty, 10 balls have to go into 2 buckets: 2^10. This also includes the cases when all the balls go to one bucket. So considering empty bucket1 also considers the cases where bucket 1 and 2 are both empty. Similarly when we consider bucket2 as empty we are also considering the case where both bucket 1 and 2 are empty. Essentially we are double subtracting the cases where 2 buckets are empty. So we need to add them back. 3C2 to choose 2 buckets to remain empty. Now each ball ...

Q1. Three distinct vertices are chosen at random from the vertices of a given regular polygon of (2n+1) sides. If all such choices are equally likely, what is the probability that the center of the given polygon lies in the interior of the triangle determined by the three chosen random points? S1. Fact: The center of a regular polygon lies strictly OUTSIDE a triangle if and only if all three chosen vertices lie on a single semicircle. Explanation: Each regular polygon has a semicircle. If there are even vertices we will have diameters joining 2 vertices. In odd, we don't. For e.g. in a pentagon, Diameter from '1' lands on side 3-4. Here 2n+1 = 5 => n = 2 Here 3 vertices are in 1 half and 2 vertices in other half. If you make a triangle using 3 consecutive vertices it won't contain the center, else it would. For e.g. 1-2-3 doesn't but 1-3-5 does. Here total ways to create a triangle 5C3 = 10 Triangles with center outside: 1-2-3, 2-3-4 ... 5-1-2 = 5 such With cente...