Functions & Calculus, Combinatorics & Induction, Probability & Random… Test for CMI Data Science: 40 Questions with Solutions & Analysis

Attempt the Functions & Calculus, Combinatorics & Induction, Probability & Random… test for CMI Data Science: 40 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

40 questions · 88 marks · 135.24 minutes. Probability Theory: 20 · School Level Mathematics: 14 · Programming: 4 · Discrete Mathematics: 2

Free sample questions from Functions & Calculus, Combinatorics & Induction, Probability & Random…

Question 1 · Probability Theory · 2025 SUB

Correct Answer:

none

Step-by-Step Solution

Insight: This is a cross-chart proportional reasoning question linking an absolute value (Lemon sales) to a percentage share to derive another absolute value (Mango sales).

Exam route: Sum Lemon bars to get absolute Lemon revenue (120 Lakhs). Divide by Lemon % (28%) to get Total Revenue. Multiply Total by Mango % (15%).

Learning route:

Step 1: Calculate total absolute revenue for Lemon flavour from the bar chart: 30 + 25 + 20 + 20 + 15 + 10 = 120 Lakhs.

Step 2: Relate Lemon revenue to Total Revenue using the pie chart. Lemon represents 28% of Total Revenue.

Step 3: Calculate Total Revenue = 120 / 0.28 = 3000 / 7 Lakhs.

Step 4: Calculate Mango revenue, which is 15% of Total Revenue: (3000 / 7) * 0.15 = 450 / 7 ≈ 64.29 Lakhs.

Question 2 · Probability Theory · 2025 SUB

Common Description: Instructions for Questions 18,19, and 20:
Read the following description carefully and answer the questions that follow. Use all the information provided. Clearly show all your calculations.
Description:
NorthCool Beverages Pvt. Ltd. is a leading beverage company that operates across six northern states of India. During the past summer the company launched an aggressive marketing campaign to promote its range of flavoured drinks. The charts below summarise the sales data collected during this campaign.
NorthCool Beverages plans to launch a new variant, “Lemon Max”, in the two states with the highest sales of the lemon flavour. The company expects Lemon Max to generate additional revenue equal to 20% of the current lemon flavour sales in those two states. What would be the percentage revenue share of lemon flavour drinks if their expectations are met?

Correct Answer:

none

Step-by-Step Solution

Insight: Adding revenue to a specific sub-category changes both the numerator (Lemon sales) and the denominator (Total sales), requiring a recalculation of the percentage share.

Exam route: Calculate total Lemon sales (120 Lakhs). Find total revenue (120 / 0.28). Identify top 2 states (UP 30, Haryana 25). Calculate additional revenue (20% of 55 = 11). New Lemon = 131, New Total = (120/0.28) + 11. Compute new share.

Learning route:

Step 1: Calculate total Lemon sales from the bar chart: 30 + 25 + 20 + 20 + 15 + 10 = 120 Lakhs.

Step 2: Calculate the initial Total Revenue. Since Lemon is 28% of Total Revenue, Total Revenue = 120 / 0.28 = 3000 / 7 ≈ 428.57 Lakhs.

Step 3: Identify the two states with the highest Lemon sales: Uttar Pradesh (30 Lakhs) and Haryana (25 Lakhs). Their sum is 55 Lakhs.

Step 4: Calculate the additional revenue from "Lemon Max": 20% of 55 = 11 Lakhs.

Step 5: Calculate the new Lemon revenue: 120 + 11 = 131 Lakhs.

Step 6: Calculate the new Total Revenue: (3000 / 7) + 11 = (3000 + 77) / 7 = 3077 / 7 Lakhs.

Step 7: Calculate the new percentage revenue share of lemon flavour drinks: (131 / (3077 / 7)) 100 = (917 / 3077) 100 ≈ 29.80%.

Question 3 · Probability Theory · 2025 SUB

Common Description: Instructions for Questions 18,19, and 20:
Read the following description carefully and answer the questions that follow. Use all the information provided. Clearly show all your calculations.
Description:
NorthCool Beverages Pvt. Ltd. is a leading beverage company that operates across six northern states of India. During the past summer the company launched an aggressive marketing campaign to promote its range of flavoured drinks. The charts below summarise the sales data collected during this campaign.
On average, for every ₹1 lakh in total revenue from the lemon-flavoured drink, 10,000 litres of lemon-flavoured drink are sold across the six states. Estimate the total volume (in litres) of lemon-flavoured drink sold by NorthCool Beverages across all six states.

Correct Answer:

none

Step-by-Step Solution

Insight: This is a direct scaling question where the total volume depends entirely on the total revenue of the specific flavour, which is already aggregated.

Exam route: Identify total Lemon revenue (120 Lakhs). Multiply by the given volume rate (10,000 litres per Lakh).

Learning route:

Step 1: Identify the total revenue for the lemon-flavoured drink from the bar chart: 30 + 25 + 20 + 20 + 15 + 10 = 120 Lakhs.

Step 2: Note the scaling factor provided: 10,000 litres per ₹1 lakh of revenue.

Step 3: Multiply the total revenue by the scaling factor: 120 * 10,000 = 1,200,000 litres.

Question 4 · School Level Mathematics · 2024 SUB

Let $α$ be a fixed real number and let $f$ be a function from $R^{2}$ to $R^{2}$ defined as

$f (x, y) = (x cos α - y sin α, x sin α + y cos α) .$

Write down an expression for $f^{10} (x, y)$ , where $f^{10}$ denotes the function obtained by composing $f$ with itself 10 times.

Correct Answer:

none

Step-by-Step Solution

Insight: The map $f$ is a geometric rotation by angle $α$ ; composing it 10 times yields a rotation by $10 α$ .

Exam route: Recognise $f = R_{α}$ . Apply $R_{α}^{n} = R_{n α}$ with $n = 10$ . Substitute $10 α$ into the rotation formula.

Learning route:

This is a function-iteration question, recognisable because the formula $(x cos α - y sin α, x sin α + y cos α)$ is the standard rotation matrix and "composing $f$ with itself 10 times" signals $f^{10} = f \circ f \circ \dots \circ f$ .

Step 1: Identify the transformation. The map $f (x, y) = (x cos α - y sin α, x sin α + y cos α)$ rotates every point anticlockwise by $α$ about the origin. We write $f = R_{α}$ .

Step 2: Iterate geometrically. One application rotates by $α$ . A second rotates the result by another $α$ , totalling $2 α$ . By induction, $n$ applications give rotation by $n α$ : $R_{α}^{n} = R_{n α}$ .

Step 3: Set $n = 10$ . Then $f^{10} = R_{10 α}$ .

Step 4: Substitute $10 α$ for $α$ in the original formula:

$f^{10} (x, y) = (x cos 10 α - y sin 10 α, x sin 10 α + y cos 10 α)$ .

Wrong path: Raising each coordinate to the 10th power, e.g. $(x cos α - y sin α)^{10}$ . This confuses iteration $f^{n}$ with exponentiation $(f (x))^{n}$ . The notation $f^{10}$ means ten-fold composition, not the tenth power of the output.

Verification: For $α = π /2$ , $f (x, y) = (- y, x)$ , so $f^{2} (x, y) = (- x, - y)$ and $f^{10} = f^{2} = (- x, - y)$ . Formula gives $(x cos 5 π - y sin 5 π, x sin 5 π + y cos 5 π) = (- x, - y)$ . Matches.

Question 5 · School Level Mathematics · 2025 SUB

Evaluate the following limit:

$x \to 0 lim \frac{e ^{2 x} - 2 x - cos x}{x sin x}$

Correct Answer:

none

Step-by-Step Solution

Insight: This is a transcendental $0/0$ limit with mixed exponential, polynomial, and trigonometric terms; the denominator's leading order $x^{2}$ dictates expanding every numerator term through $x^{2}$ using Taylor series.

Exam route:

Denominator: $x sin x = x (x - x^{3} /6 + \dots) = x^{2} + O (x^{4}) \sim x^{2}$ .
Numerator to $O (x^{2})$ :

$e^{2 x} = 1 + 2 x + 2 x^{2} + O (x^{3})$ ,

$- 2 x = - 2 x$ ,

$- cos x = - (1 - x^{2} /2 + O (x^{4})) = - 1 + x^{2} /2 + O (x^{4})$ .

Assemble: $(1 + 2 x + 2 x^{2}) - 2 x + (- 1 + x^{2} /2) = \frac{5}{2} x^{2} + O (x^{3})$ .
Divide: $lim_{x \to 0} \frac{\frac{5}{2} x ^{2} + O ( x ^{3} )}{x ^{2} + O ( x ^{4} )} = \frac{5}{2}$ .

Learning route:

This is a transcendental $0/0$ limit, recognisable because direct substitution gives $(1 - 0 - 1) / (0 \cdot 0) = 0/0$ and the numerator mixes exponential, polynomial, and trigonometric terms. The trigger for Taylor expansion is this mixture — algebraic factoring cannot untangle $e^{2 x}$ and $cos x$ simultaneously.

Step 1: Determine the denominator's leading order. Expand $sin x = x - x^{3} /6 + \dots$ , so:

$x sin x = x (x - \frac{x ^{3}}{6} + \dots) = x^{2} - \frac{x ^{4}}{6} + \dots = x^{2} + O (x^{4})$

The leading term is $x^{2}$ . This tells us we must expand every numerator term through $x^{2}$ — no less, no more.

Step 2: Expand each numerator term to $O (x^{2})$ :

$e^{2 x} = 1 + (2 x) + \frac{( 2 x ) ^{2}}{2 !} + O (x^{3}) = 1 + 2 x + 2 x^{2} + O (x^{3})$

$- 2 x = - 2 x$

$- cos x = - (1 - \frac{x ^{2}}{2} + O (x^{4})) = - 1 + \frac{x ^{2}}{2} + O (x^{4})$

Step 3: Combine all numerator terms:

$(1 + 2 x + 2 x^{2}) - 2 x + (- 1 + \frac{x ^{2}}{2}) = \frac{5}{2} x^{2} + O (x^{3})$

Step 4: Divide by the denominator and take the limit:

$x \to 0 lim \frac{\frac{5}{2} x ^{2} + O ( x ^{3} )}{x ^{2} + O ( x ^{4} )} = x \to 0 lim \frac{\frac{5}{2} + O ( x )}{1 + O ( x ^{2} )} = \frac{5}{2}$

Wrong path: Using L'Hôpital's rule twice is possible but extremely messy and prone to calculation errors. Another common mistake is expanding the numerator only to $O (x)$ , which yields $0$ in the numerator and leads to the incorrect conclusion that the limit is $0$ .

Question 6 · School Level Mathematics SUB

Evaluate

$x \to 0 lim \frac{e ^{x} - 1 - sin x}{x ^{2}} .$

Correct Answer:

0.5

Step-by-Step Solution

Key idea: This is a mixed transcendental $0/0$ limit combining exponential and trigonometric functions. The denominator is $x^{2}$ , so we need Taylor expansions up to order $x^{2}$ for every term in the numerator.

Step 1: Verify the indeterminate form. At $x = 0$ : $e^{0} - 1 - sin 0 = 1 - 1 - 0 = 0$ . Denominator is $0$ . Confirmed $0/0$ .

Step 2: Write Taylor expansions to order $x^{2}$ .

$e^{x} = 1 + x + \frac{x ^{2}}{2} + O (x^{3})$

$sin x = x - \frac{x ^{3}}{6} + O (x^{5}) = x + O (x^{3})$ (we only need up to $x^{2}$ )

Step 3: Substitute into the numerator.

$e^{x} - 1 - sin x = (1 + x + \frac{x ^{2}}{2} + O (x^{3})) - 1 - (x + O (x^{3}))$

$= 1 + x + \frac{x ^{2}}{2} - 1 - x + O (x^{3})$

$= \frac{x ^{2}}{2} + O (x^{3})$

Step 4: Divide by the denominator.

$\frac{\frac{x ^{2}}{2} + O ( x ^{3} )}{x ^{2}} = \frac{1}{2} + O (x)$

Step 5: Take the limit as $x \to 0$ .

The $O (x)$ terms vanish, leaving $\frac{1}{2}$ .

Answer: 0.5

Question 7 · Programming · 2021 SUB

Consider the following code where

A

is an array indexed from 0.
function foo(A, year, n) {
    l = 0, r = n - 1, c = 0;
    while (l <= r) {
        c = c + 1;
        m = l + (r - l) // 2;
        if (A[m] == year) {
            return(c * m);
        }
        if (A[m] < year) {
            l = m + 1;
        } else {
            r = m - 1;
        }
    }
    return(-1);
}

function bar() {
    A = [2016, 2017, 2018, 2019, 2020, 2021, 2022];
    result = foo(A, 2021, 7);
    print(result);
}

Here,

a // b

represents integer division. For example,

23//10 = 2

. What will be printed when bar() is executed?

Correct Answer:

Step-by-Step Solution

Key idea: Trace the binary search algorithm to count the number of iterations ( $c$ ) and find the index ( $m$ ) where the target is found. Then compute $c \times m$ .

Step 1: Initialize variables.

$A = [2016, 2017, 2018, 2019, 2020, 2021, 2022]$

$n = 7$

$y e a r = 2021$

$l = 0, r = 6, c = 0$

Step 2: Iteration 1.

Condition $l \leq r$ ( $0 \leq 6$ ) is True.

$c = c + 1 = 1$ .

$m = l + (r - l) //2 = 0 + (6 - 0) //2 = 3$ .

Check $A [3] == 2021$ ? $A [3] = 2019$ . $2019 \neq = 2021$ .

Check $A [3] < y e a r$ ? $2019 < 2021$ is True.

Update $l = m + 1 = 4$ .

( $r$ remains 6).

Step 3: Iteration 2.

Condition $l \leq r$ ( $4 \leq 6$ ) is True.

$c = c + 1 = 2$ .

$m = l + (r - l) //2 = 4 + (6 - 4) //2 = 4 + 1 = 5$ .

Check $A [5] == 2021$ ? $A [5] = 2021$ . Yes!

Return $c \times m = 2 \times 5 = 10$ .

Wait, let me re-read the code carefully.

m = l + (r - l) // 2

Re-trace Iteration 1:

$l = 0, r = 6$ .

$m = 0 + (6//2) = 3$ .

$A [3] = 2019$ .

$2019 < 2021$ . True.

$l = 3 + 1 = 4$ .

$c = 1$ .

Re-trace Iteration 2:

$l = 4, r = 6$ .

$m = 4 + (6 - 4) //2 = 4 + 1 = 5$ .

$A [5] = 2021$ .

$A [5] == y e a r$ . True.

Return $c \times m$ .

Current $c$ is incremented at start of loop.

Start of Iter 2: $c$ becomes 2.

So return $2 \times 5 = 10$ .

Let me double check the array indices.

Index: 0: 2016

1: 2017

2: 2018

3: 2019

4: 2020

5: 2021

6: 2022

Target 2021 is at index 5.

Trace again.

Init: l=0, r=6, c=0.

Loop 1:

l<=r (0<=6) True.

c=1.

m = 0 + (6-0)//2 = 3.

A[3] = 2019.

A[3] == 2021? No.

A[3] < 2021? Yes.

l = 3+1 = 4.

Loop 2:

l<=r (4<=6) True.

c=2.

m = 4 + (6-4)//2 = 4 + 1 = 5.

A[5] = 2021.

A[5] == 2021? Yes.

Return c m = 2 5 = 10.

Answer: 10.

Question 8 · Programming MSQ

Consider the standard bubble sort algorithm applied to an array of 5 distinct integers. How many possible initial permutations of this array will result in exactly 3 swaps being performed during the entire sorting process?

A.
6
B.
10
C.
15
D.
20

Correct Answer:

Step-by-Step Solution

Key idea: Recognize that the total number of swaps in bubble sort is exactly equal to the number of inversions in the initial array, then count the permutations with exactly 3 inversions.

Step 1: Establish the invariant. Each swap in bubble sort exchanges two adjacent elements that are out of order. This operation reduces the total number of inversions in the array by exactly 1. The algorithm terminates when the array is sorted (0 inversions). Therefore, Total Swaps = Initial Inversions.

Step 2: Reframe the problem. We need to find the number of permutations of 5 distinct elements that have exactly 3 inversions.

Step 3: Use the inversion sequence representation. Any permutation of $n$ elements can be uniquely represented by an inversion sequence $(c_{1}, c_{2}, \dots, c_{n})$ , where $c_{i}$ is the number of elements to the left of the $i$ -th element that are greater than it. The constraints are $0 \leq c_{i} \leq i - 1$ .

Step 4: We need the number of sequences $(c_{1}, c_{2}, c_{3}, c_{4}, c_{5})$ such that $\sum_{i = 1}^{5} c_{i} = 3$ , with $c_{1} = 0$ , $c_{2} \leq 1$ , $c_{3} \leq 2$ , $c_{4} \leq 3$ , $c_{5} \leq 4$ .

Step 5: Enumerate the valid sequences for $c_{2} + c_{3} + c_{4} + c_{5} = 3$ :

$c_{5} = 3$ : (0, 0, 0, 3) $\to$ 1 way
$c_{5} = 2$ : (0, 0, 1, 2), (0, 1, 0, 2), (1, 0, 0, 2) $\to$ 3 ways
$c_{5} = 1$ : (0, 0, 2, 1), (0, 1, 1, 1), (0, 2, 0, 1), (1, 0, 1, 1), (1, 1, 0, 1) $\to$ 5 ways
$c_{5} = 0$ : (0, 0, 3, 0), (0, 1, 2, 0), (0, 2, 1, 0), (1, 0, 2, 0), (1, 1, 1, 0), (1, 2, 0, 0) $\to$ 6 ways

Step 6: Sum the possibilities: $1 + 3 + 5 + 6 = 15$ .

Answer: 15

Question 9 · Programming MSQ

Consider the optimized bubble sort algorithm (which includes a boolean flag to terminate early if a pass completes with zero swaps). How many permutations of the array ${1, 2, 3, 4}$ will require exactly 2 passes of the outer loop to fully sort the array?

A.
6
B.
7
C.
8
D.
9

Correct Answer:

Step-by-Step Solution

Key idea: Relate the number of passes in optimized bubble sort to the maximum inversion count of any single element, then bound and count the valid permutations.

Step 1: Establish the invariant. In optimized bubble sort, the algorithm terminates after $k$ passes if and only if the maximum number of inversions for any single element in the initial array is $k - 1$ .

Step 2: Reframe the problem. We need to find the number of permutations of ${1, 2, 3, 4}$ where the maximum inversion count for any element is exactly $1$ . (If it were 0, it would take 1 pass; if $\geq 2$ , it would take $\geq 3$ passes).

Step 3: Use the inversion sequence representation. A permutation can be uniquely represented by $(c_{1}, c_{2}, c_{3}, c_{4})$ , where $c_{i}$ is the number of elements to the left of the $i$ -th smallest element that are greater than it. The constraints are $0 \leq c_{i} \leq i - 1$ .

Step 4: We require $max (c_{1}, c_{2}, c_{3}, c_{4}) \leq 1$ .

Since $c_{1} = 0$ always, we need $c_{2} \in {0, 1}$ , $c_{3} \in {0, 1}$ , and $c_{4} \in {0, 1}$ .

Step 5: Count the valid sequences. There are $1 \times 2 \times 2 \times 2 = 8$ such sequences. Each uniquely maps to a valid permutation.

Step 6: Exclude the case requiring only 1 pass. The sequence $(0, 0, 0, 0)$ corresponds to the already sorted array $[1, 2, 3, 4]$ , which requires exactly 1 pass (0 swaps in the first pass).

Step 7: Calculate the final count. The number of permutations requiring exactly 2 passes is $8 - 1 = 7$ .

Answer: 7

Question 10 · Discrete Mathematics · 2023 SUB

Consider a city with

n

East-West Streets (EWS) and

n

North-South Avenues (NSA). The EWS are the line segments

{y = j ∣ 1 \leq x \leq n}

for all

j \in {1, 2, \dots, n}

and the NSA are the line segments

{x = i ∣ 1 \leq y \leq n}

for all

i \in {1, 2, \dots, n}

. A junction is a pair

(i, j)

where avenue

i

intersects street

j

. How many junctions are there in the city?
For increased safety, the city council decides to place cameras at various junctions in the city. The cameras being super-powerful, can observe the entire street and avenue corresponding to the junction that they are placed at. For instance, a camera placed at the junction

(i, j)

can observe both avenue

i

and street

j

. Write down an expression for the minimum number of cameras needed to observe every street and avenue in the city.
Justify your answers.

Correct Answer:

none

Step-by-Step Solution

Insight: The streets and avenues are explicitly $n$ lines each (not cell boundaries of an $n \times n$ board), so junctions $= n^{2}$ ; the camera problem is a minimum edge-cover on $K_{n, n}$ , which equals $n$ by the diagonal construction.

Exam route:

Part 1 — The EWS are $y = 1, y = 2, \dots, y = n$ : exactly $n$ horizontal line segments. The NSA are $x = 1, x = 2, \dots, x = n$ : exactly $n$ vertical line segments. A junction is the intersection of one avenue with one street, so by the product rule the number of junctions is $n \times n = n^{2}$ .

Part 2 — A camera at $(i, j)$ covers avenue $i$ and street $j$ , i.e. exactly one avenue and one street. To cover all $n$ avenues we need at least $n$ cameras (each camera covers at most one avenue). The diagonal placement $(1, 1), (2, 2), \dots, (n, n)$ uses exactly $n$ cameras and covers every avenue $i$ (via camera $(i, i)$ ) and every street $j$ (via camera $(j, j)$ ). Therefore the minimum number of cameras is $n$ .

Learning route:

This is a grid-enumeration plus covering-argument question, recognisable because the problem defines lines by explicit equations and then asks for a minimum set of junctions that "dominate" all lines.

Step 1 — Count the lines carefully. The set ${y = j ∣ 1 \leq x \leq n}$ for $j \in {1, 2, \dots, n}$ contains exactly $n$ distinct horizontal line segments. Similarly the avenues ${x = i ∣ 1 \leq y \leq n}$ for $i \in {1, 2, \dots, n}$ give exactly $n$ vertical segments. Crucially, the problem gives us $n$ lines in each direction, NOT $n$ cells. This is the opposite of the usual chessboard setup where an $n \times n$ board has $n + 1$ lines.

Step 2 — Count junctions. Each junction is uniquely determined by choosing one avenue and one street. Since there are $n$ choices for the avenue and $n$ choices for the street, the total number of junctions is $n \times n = n^{2}$ .

Step 3 — Model the camera problem. A camera at junction $(i, j)$ observes avenue $i$ and street $j$ . We need every avenue and every street to be observed by at least one camera. This is equivalent to finding a minimum edge cover in the complete bipartite graph $K_{n, n}$ where left vertices = avenues, right vertices = streets, and edges = junctions.

Step 4 — Lower bound. Each camera covers exactly one avenue. Since there are $n$ avenues to cover, we need at least $n$ cameras.

Step 5 — Construction. Place cameras at $(1, 1), (2, 2), \dots, (n, n)$ . Camera $(k, k)$ covers avenue $k$ and street $k$ . As $k$ ranges from $1$ to $n$ , all $n$ avenues and all $n$ streets are covered. This uses exactly $n$ cameras, matching the lower bound.

Verification: For $n = 3$ , we have 3 streets and 3 avenues, giving $3^{2} = 9$ junctions. Cameras at $(1, 1), (2, 2), (3, 3)$ cover all 3 avenues and all 3 streets. No set of 2 cameras can cover 3 avenues (each covers at most 1). So the answer $n = 3$ is correct.

Question 11 · Discrete Mathematics MSQ

A bug travels from the bottom-left junction $(0, 0)$ to the top-right junction $(5, 5)$ of a $5 \times 5$ grid of cells, moving only one unit right or one unit up at each step. How many such shortest paths do not pass through the junction $(2, 2)$ or the junction $(3, 3)$ ?

A.
84
B.
132
C.
72
D.
252

Correct Answer:

Step-by-Step Solution

Key idea: This is a shortest path counting problem with multiple forbidden points, best solved using the Principle of Inclusion-Exclusion (PIE).

Step 1: Calculate the total number of unrestricted shortest paths from $(0, 0)$ to $(5, 5)$ . This requires 5 right and 5 up moves, totaling 10 moves. The number of ways is $(5 10) = 252$ .

Step 2: Calculate paths passing through $(2, 2)$ . This is paths from $(0, 0)$ to $(2, 2)$ multiplied by paths from $(2, 2)$ to $(5, 5)$ .

$(2 4) \times (3 6) = 6 \times 20 = 120$ .

Step 3: Calculate paths passing through $(3, 3)$ . This is paths from $(0, 0)$ to $(3, 3)$ multiplied by paths from $(3, 3)$ to $(5, 5)$ .

$(3 6) \times (2 4) = 20 \times 6 = 120$ .

Step 4: Calculate paths passing through both $(2, 2)$ and $(3, 3)$ . This requires going $(0, 0) \to (2, 2) \to (3, 3) \to (5, 5)$ .

$(2 4) \times (1 2) \times (2 4) = 6 \times 2 \times 6 = 72$ .

Step 5: Apply PIE. Valid paths = Total - (Through 2,2) - (Through 3,3) + (Through both).

Valid paths = $252 - 120 - 120 + 72 = 84$ .

Answer: 84.

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Functions & Calculus, Combinatorics & Induction, Probability & Random… Test for CMI Data Science: 40 Questions with Solutions & Analysis

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