Quant Boosters - Soumya Chakraborty - Set 1
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
Q26) 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:
- For 10 intervals (12th to 22nd term), we move 40 up, so for one interval forward, we should move 4 up.
- 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
Q27) 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)
Q28) How many 4 digit numbers divisible by 6 can be formed using the digits 0,1,2,3,4,5 when repetition is allowed?
Let us understand the constraints first:
- The leftmost digit cannot be zero
- The last digit needs to be even
- 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
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
Q29) 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
Q30) 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:
- All distinct number of balls
- Two identical number of balls
- 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
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