Math IOQM

 

 

1. Prove that if p/q is a rational number in its lowest form then it will terminate if and only if denominator = 2^m.5^n and m,n are whole numbers. And the number of decimals will be max(m,n). 2. Prove sqrt(2) is irrational. (Assume it isn't and then p/q and p,q are both even). 3. a < (a+b)/2 < b for all real a,b 4. Fact: A rational number in decimal form will either terminate or start repeating after some digits. It's not possible for a rational to never terminate and not cycle at the same time. 5. Fact: Pie = 4 * (1/1 - 1/3 + 1/5 - 1/7 + 1/9 ...) Basically alternating odd denominators.

Q1. A DNA chain is composed of basic building blocks in the form of four chemicals, known by the symbols A,C,T,G. Consider three-letter chains consisting of these symbols (with or without repetitions). How many such chains are there? S1. 4*4*4 = 64 Q2. Consider the set {a,b,c,d,e}. How many three-letter words can be made out of them, with or without meaning? How many of these will have at least one vowel in them? Answer these questions both for the case when repetitions of letters are allowed and for the case when repetitions of letters are not allowed. S2. 2.1 With repetition: 5*5*5 = 125 Without: 5*4*3 = 60 2.2 With repetition: Total words: 5^3 = 125 Without any vowels = 3^3 = 27 Answer = 125 - 27 = 98 Without: Total: 5*4*3 = 60 Without vowels = 3*2*1 Answer = 60 - 6 = 54 Q3. A permutation (a1,a2,a3,a4,a5) of (1,2,3,4,5) is called heavy-tailed if a1 + a2 < a4 + a5. What is the number of heavy-tailed permutations? S3. Total such: 5! = 120 Let's fix 1,2 to a1,a2 => 2!*3! = 12...

1. Find the total number of words formed by the letters a,a,b,b,c. S1. 5!/2!.2! 2. In how many ways can you arrange 2 beads of red colour, 2 beads of blue colour and 3 beads of black colour in a ring? Note that beads with the same colour are identical and arrangements that can be obtained by rotating (or) flipping the ring are identical. S2. First attempt(wrong): (7-1)! circular permutations Divide by 2!2!3! to account for repeated elements. Divide by 2 to account for flipping. Answer = 15 This is wrong, correct answer is 18. Why? Let's see using a smaller example. 2 blue, 3 black. (5-1)! circular permutations Divide by 2!3!. So far so good. Mistake is when we divide by 2 for flipping. Why do we even divide by 2 for flipping. Let's say we have 4 distinct beads. Here is one necklace: 1 --- 2 |       | 4 -- 3 and flipping it we get: 2 --- 1 |       | 3 -- 4 So 1-2-3-4 was clockwise and it became counter clockwise. Flipping gives a distinct perm...

1. 5 chairs, 4 people? How many ways can they sit? 5C4*4! = 120 2. 10 mangoes, 12 apples, 16 guava, 15 bananas. How many ways to eat 3 fruits? Case 1: Every single fruit is distinct. We have 53 fruits in total. So 53C3. That is if the order of eating doesn't matter. If the order of eating matters, then: 53.52.51 Case 2: Fruits of one type are all same.  Case 2.1 - Order of eating doesn't matter. Then this is simply stars and bars. (n+k-1)C(n) where 'n' is the number of items and 'k' is the number of buckets. Here,  n = 3 k = 4 So: All 3 fruits are same: 4C1 = 4 2 Same 1 distinct: 4C2 * 2C1 = 12 All 3 distinct: 4C3 = 4 Total: 20 Case 2.2 - Order of eating does matter. 4 * 4 * 4 = 64 Case 3: One fruit of each type, i.e. I eat 1 mango, 1 apple, 1 banana for e.g. Case 3.1 Fruits of same type are also distinct We have to consider 4 different cases(4C3): Mango, banana, apple = 10*15*12 Mango, banana, guava = 10*15*16 Mango, apple, guava = 10 * 12 * 16 Apple, banana, g...