In a certain test, a_i students gave...

MathematicsFundamental Principle of CountingJEE Advanced 1982Moderate

In a certain test, $a_{i}$ students gave wrong answers to atleast $i$ questions, where $i = 1, 2, \dots, k$ . No student gave more than $k$ wrong answers. The total number of wrong answers given is .........

Answer:

a_{1} + a_{2} + \dots + a_{k}

Visualized Solution (English)

Defining the Variable .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n_{i}

Let $n_{i}$ be the number of students who gave exactly $i$ wrong answers.
Here, $i$ can range from $1, 2, \dots, k$ .
Note that no student gave more than $k$ wrong answers, so $n_{i} = 0$ for $i > k$ .

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a_{i} to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n_{i}

The given value $a_{i}$ is the number of students with at least $i$ wrong answers.
Mathematically: $a_{i} = n_{i} + n_{i + 1} + n_{i + 2} + \dots + n_{k}$ .
This means $a_{1}$ is the sum of all students who made any mistakes.

The Total Mistakes Formula

Total number of wrong answers $W$ is the sum of mistakes made by each student.
$W = 1 \cdot n_{1} + 2 \cdot n_{2} + 3 \cdot n_{3} + \dots + k \cdot n_{k}$ .
In summation notation: $W = \sum_{i = 1}^{k} i \cdot n_{i}$ .

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a_{i} - a_{i + 1} Relationship

From our definition: $a_{i} = n_{i} + n_{i + 1} + \dots + n_{k}$ and $a_{i + 1} = n_{i + 1} + \dots + n_{k}$ .
Subtracting these gives: $n_{i} = a_{i} - a_{i + 1}$ for $i = 1, 2, \dots, k - 1$ .
For the last term: $n_{k} = a_{k}$ (since no student has more than $k$ mistakes).

Substituting and Expanding

Substitute $n_{i}$ into the total sum $W$ :
$W = 1 (a_{1} - a_{2}) + 2 (a_{2} - a_{3}) + 3 (a_{3} - a_{4}) + \dots + (k - 1) (a_{k - 1} - a_{k}) + k \cdot a_{k}$ .

Simplifying the Sum

Rearrange the terms by grouping $a_{i}$ :
$W = 1 \cdot a_{1} + (- 1 + 2) a_{2} + (- 2 + 3) a_{3} + \dots + (- (k - 1) + k) a_{k}$
$W = a_{1} + a_{2} + a_{3} + \dots + a_{k}$ .

The Way Forward

Key Takeaway: The total count of items can be found by summing the 'at least $i$ ' counts.
This is a discrete version of the identity $\int P (X > x) d x = E [X]$ .
Next Challenge: What if the question asked for the sum of squares of the number of wrong answers? How would the formula change?

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