MathOratory - Quant notes - Set 3


  • IIM Lucknow | MathOratory


    In how many ways can one place 20 different books in 5 shelves assuming that the books can fit into any one of them all at once?

    The other answer I saw is the same way that I used to explain in my class as well. But, there's an alternative way to reach the answer as well

    First thing that we should know is that if there are 'n' units kept in a row, then there are 'n+1' gaps generated by them. So, what we should understand is that when there is no item there is 1 gap, when there is one item there are 2 gaps, and so on.

    Let us use that concept here, with a small variation: the 5 shelves are 5 initial gaps. The first book has 5 choices (5 initial gaps) to go to. Once we have placed this book, there are now 6 gaps on which the 2nd book can be placed. After the 2nd book is placed, there are 7 gaps in which 3rd book can be placed, and so on.

    So, the total arrangements will be = 5 * 6 * 7 *... 24 [as there are 20 books] = 24P20

    For what value of a positive integer n does the LCM of n and 36 exceed their HCF by 500?

    If we take the two numbers as h.a and h.b, where the HCF is 'h'. Then the LCM must be h.a.b, 'a' and 'b' must be co-prime to each other.

    The difference between LCM and HCF will be h.(a.b - 1) which would (let's call these criteria)

    1. Also have a factor of 'h'
    2. The HCF of either of the numbers and difference between LCM and HCF would also be 'h'

    We have two critical information in the question (let's call these data)

    36 is one of the numbers
    500 is the difference between LCM and HCF
    The HCF of 36 and 500 being 4, using 2nd criteria, the HCF of the two numbers must also be 4. One of the numbers being 4 * 9, the other must be 4 * a.

    LCM would be 4ab. Hence, 4 * 9 * a - 4 = 500
    a = 14
    So, the other number is 4 * 14 = 56

    What is the smallest 5 digit palindrome divisible by 45?

    Firstly, divisible by 45 implies divisible by 5&9

    In order to be divisible by 5, the last digit has to be 0 or 5, in this case 0 making no sense, as the number is palindromic and the last digit has to be same as the leftmost digit, which can't be 0. So the starting and the last digit needs to be 5, the number must be of the form 5ABA5

    Now, it must also be divisible by 9, implying the sum of digits must be divisible by 9. 5+5=10, which divided by 9, leaves a remainder of 1, so the remaining three digits must leave a remainder of 8, when added. As, we need to minimise the number we should take A+B+A = 8

    Ideally we maximise digits from the right side, if the sum is constant and the number has to be minimised. But in this case the 2nd and the 4th digit are same. So, we must try to minimise the 2nd digit from the left, in order to minimise the number. So, assign 8 to the third digit, and we have the minimum number as 50805

    If a bag contains five balls. Two balls are drawn and are found to be white. What is the probability that all the balls are white?

    It's ideally a Bayes' theorem application, but why complicate matters, it can be (like all other probability questions) solved quite logically.

    First, we have to understand what are the information and what is the objective ?

    We don't know the number of white and black balls in the bag, but we know the number of white balls can't be less than 2.

    Now, there are different cases, which are the number of white balls in the bag. The total cases are : 2c2+3c2+4c2+5c2 =20

    The question is asking to find the probability of the case that there are 5 white balls, which 5c2 = 10 cases

    So, probability =10/20=1/2

    What is total number of positive integral solutions for (x, y, z) such that xyz=24?

    There's a strong formula, famously known as the stars and bars algorithm, which simplifies this solution for any product.

    I'll just detail the solution once I've explained the formula. It's used, ideally, for counting the number of ways of distributing 'n' identical balls into 'r' different boxes. The number of ways of doing that is (n+r-1)C(r-1). The formula can be easily remembered but hardly the applications understood. Let's understand what is the primary difference between distributing 'n' IDENTICAL balls and 'n' DISTINCT balls. The idea behind DISTINCT balls is that in this case, not only how many balls go into a particular box matter, but also WHICH ones matter. But when they are IDENTICAL balls, only thing that matters is how many goes into a specific box.

    Here XYZ =24 = 2^3 * 3

    So, we have three identical 2s to distribute into three boxes (X,Y,Z). The number of ways of doing that would be (3+3-1)C(3-1) = 5C2 = 10 ways.

    Correspondingly, the number of ways of distributing the 3 into the three boxes must be 3 ways

    So, we have a total of 10*3=30 ways of distributing

    An AP consists of 23 terms. If the sum of the three terms in the middle is 141 and the sum of the last three terms is 261,then the first term is?

    I like equations, but I like numbers more. So, I'm going to solve this numerically.

    The avg of the 11,12,13th terms will be the 12th term. So, the 12th term would be 141/3 = 47

    The avg of the 21,22,23rd terms will be the 22nd term. So, the 22nd term would be 261/3 = 87

    So, couple of things:

    1. For 10 intervals (12th to 22nd term), we move 40 up, so for one interval forward, we should move 4 up.
    2. The 22nd term is 40 more than 12th. So, 2nd term must be 40 less than the 12th.

    So, 2nd term = 47-40=7
    And the first term would be 7-4=3

    How many combinations of 2 or more consecutive positive whole numbers (integers) when added together will equal the sum of 50?

    In this case, simply put, we are dealing with an AP series with a common difference of 1.

    Now, for an AP series with odd number of terms, the average is the middle term. And for an AP series with even number of terms, the average is halfway between the middle two terms. *There's a small constraint, that the number of terms should be at least two.

    If, the number of terms is odd, the middle term (being the average) must be equal to 50/n, and as the middle term is a term of the series, must be integer. So, if 'n' is odd then 50 must be divisible by n.

    If, the number of terms is even, the halfway point between the middle terms must be equal to 50/n. As, the middle terms are consecutive integers, their halfway point should be of the form 'x.5'

    So, if 'n' is odd it must be factor of 2 * 5^2 (=50), or in other words, n must be a factor of 5^2. And, if 'n' is even then when 2 * 5^2 is divided by n, then we should have something of the form x.5, or in other words n must have 2^2 multiplied to some factor of 5^2.

    Before we begin the cases, the minimum number in the series must be positive

    Case1: n is odd

    N=5, then the average is 50/5=10.Now, as the middle term is the 3rd in the series, if we move 2 terms back, we have a positive value, 10−2

    N=25, then the average is 50/25=2..But in this case if we move 12 terms back, we have the minimum as negative

    So only 1 case for n being odd

    Case2: n is even

    N = 2^2∗1, then the average is50/4=12.5, which makes the middle two terms 12 and 13. As, there are 4 terms only, moving one term back from 12 will keep the minimum value positive.

    N = 2^2∗5, then the average is 50/20=2.5. wherein we have 20 terms. So, moving 10 terms back from 2, will make the minimum negative.

    We don't have to look for the next cases, by increasing the values of N. As, in further cases, the average will decrease and with more to go back to the minimum, will surely lead to negative

    So this also have 1 case

    So, we have a total of 2 cases (Ans)

    How many 4 digit numbers divisible by 6 can be formed using the digits 0,1,2,3,4,5 when repetition is allowed?

    Step-by-step approach
    Let us understand the constraints first:

    1. The leftmost digit cannot be zero
    2. The last digit needs to be even
    3. The sum of digits needs to be even

    Now, let us understand firstly how to use the constraints in PnC (this would be more effective for the case of repetition not being allowed)
    The first constraint is not something we must be beginning with, in other words we may or may not
    The second constraint is something which limits the cases to half
    The third one is basically a constraint that is focused on the last choice we take. The last choice will be dependent on the sum of the previous three choices.

    So let us select the digits in that order
    Second constraint limits the last place to 3 choices
    The first constraint limits the 1st place to 5 choices (repetition allowed)
    ps: see that in repetition being allowed, the 1st two constraints are independent
    The third one says that the last one is dependent on the sum of the first three, so 2nd place we can choose from any 6 options. But, then the next one whatever be will have 2 choices. explained below
    So total cases would be 3 * 5 * 6 * 2 = 180 cases

    if till the third digit that we select, the sum is of the form 3k, then the fourth value should be of the form 3k (which are 2 possibilities here, viz. 0&3)
    if till the third digit ..........of the form 3k+1, then the fourth value should be of the form 3k+2 (which are also 2 possibilities)
    I think by now you should understand that there are 6 consecutive possibilities, and 6 is a multiple of 3

    Logical/Observational approach

    Firstly, you should know that this process would work only when repetition is allowed and few other constraints.

    The lucky part is that the number of digits that we can use is actually 6. And more so, they are consecutive. So, here we can employ the third constraint of the previous solution and merge it up with 2nd.
    So, for a four digit number ABCD, we are focusing on D
    For a specific set of ABC, we would have 6 consecutive numbers
    ABC0 to ABC5 and (number theory)/(logic) tells us that out of 6 consecutive numbers exactly 1 must be divisible by 6

    So, our random chance of selection would be 1/6 of the total cases

    Thus the answer should be (1/6) * 5 * 6 * 6 * 6 =180 cases

    In how many ways can you express 72 as the product of 3 natural numbers (unordered pairs)?

    Let us factorise 72 first, that'll be 2^3*3^2

    Now, a * b * c =2^3 * 3^2

    The number of ways the 3 powers of 2 is distributed over the three variables would be (3+3-1)c(3-1) =5c2 = 10 ways
    The number of ways the 2 powers of 3 is distributed over the three variables would be (2+3-1)c(3-1) =4c2 = 6 ways

    Now, these are corresponding ways
    So, total ways= 10*6=60 ways

    Now, to deal with the unordered part
    The number of solutions in which 72 is represented as the product of two identical numbers and a unique one, will be decided by the number of perfect squares that are factors of 72 (i.e. a * a * b = 72)

    Number of perfect square factors of 72 = (1+1)*(1+1) = 4

    Thus, here are 4 cases of triplets of the form a,a,b
    Now, these would have been counted 3!/2!=3 times in ordered triplets, so, we have 4*3=12 such cases

    The remaining 60-12 = 48 cases must have all distinct values (all identical are not possible)
    These 48 cases have each distinct unordered triplets counted 3!=6 times. Thus there are 48/6=8 triplets which have all distinct values and 4 triplets in which there are two identical.

    Total unordered triplets = 8 + 4 = 12

    Imagine 12 identical balls that are placed in 3 identical boxes. What is the probability that one of the boxes contains exactly 3 balls?

    The total number of ways of distributing 12 identical balls into 3 DISTINCT boxes would be (12+3-1)c(3-1) = 91

    Now, as the boxes are identical, what would be the difference? If I have a distribution of 1,2,9 balls in the boxes, then this specific distribution can be done 3!=6 ways when the boxes are distinct, but in only ONE way, if they are identical.
    Similarly, if we have a distribution of the type 2,2,8 balls, then this can be done in 3!/2!=3 ways in distinct boxes, but again in only ONE way for identical boxes.
    If, we have the distribution 4,4,4 balls, then that can be done in only ONE way in both the cases.

    So, first I'll distribute the 91 cases into three types:

    1. All distinct number of balls
    2. Two identical number of balls
    3. All identical number of balls

    What we are going to do is, we'll begin with type 2. In this scenario, we have let's say a,a,b balls in the boxes.
    We have 2a+b=12, whose number of solutions is 6+1=7
    But out of those 7 cases, 1 is the distribution 4,4,4 which does not fall under this scenario.

    So, we have 6 cases under type 3, each of which has been counted thrice, and 1 case of type 3, which has been counted once. And all other are of type 1 which has counted 6 times

    Hence distinct possibilities are = 6/3 + 1 + (91-7)/6 = 17
    This would determine all possible outcomes.

    Now, it's easy to find the favourable outcomes
    3+a+b=12
    Number of solutions for (a,b) will be (0,9); (1,8) ... (4,5): which is a total of 5 cases

    So, probability = 5/17


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