Sigma Percentile
JEE Main 2025 (January)
LEVELBoard

Animated Solution for Mathematics - Permutations and Combinations: Number of functions , that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to

Enter Numerical Value:

Visualized Solution

Defining the Sets

  • Domain:
  • Codomain:
  • We need to find the number of functions .

Splitting the Domain

  • The constraint applies only to the first elements.
  • Let's split the domain into two subsets:

Constraint on Subset

  • Exactly one element must satisfy .
  • Let this element be .
  • All other elements in must map to .

Calculating Ways for

  • Number of ways to choose the element is .
  • Ways for .
  • The remaining elements have only choice each.

Analyzing Subset

  • Subset .
  • There are no constraints on these elements.
  • Element can map to or ( choices).
  • Element can map to or ( choices).

Calculating Ways for

  • Total ways to map elements of :
  • ways.

Total Number of Functions

  • Total functions = (Ways for ) (Ways for )
  • Total functions =
  • Total functions =

The Sigma Insight: Fundamental Principle of Counting

Solution Diagram

The Art of Counting

A Journey Through Functions
Welcome, fellow traveler in the world of mathematics! Today, we are going to unravel a beautiful problem in combinatorics. It is not just about finding a number; it is about understanding the structure of functions.
Imagine you are standing before a set of 100 items, and you have to assign each of them a value of either 0 or 1. There is a constraint: exactly one element from the set must map to 1. Let us break this down together.

Analyzing the Setup

When a problem seems large, the first instinct of a mathematician is to simplify. We have a domain and a codomain .
The constraint is local; it only cares about the first 98 integers. So, let us partition our domain into two distinct subsets:
1. 2.
By separating the constrained elements from the free ones, we turn one complex problem into two manageable ones. This is the essence of combinatorial thinking.

The Special One in

Now, let us focus on . The problem demands that exactly one element in this set maps to 1.
Think of this as choosing a 'special' element. If we have 98 elements, the number of ways to choose one to be the special one is given by the combination formula:
This gives us 98 ways to pick the element that maps to 1. Once that element is chosen, all other 97 elements in are forced to map to 0. They have no choice, so the number of ways to map is exactly 98.

The Freedom of

Now, look at . The problem is silent about these two, meaning they are free agents.
Element 99 can map to 0 or 1, providing 2 choices. Element 100 can also map to 0 or 1, providing another 2 choices.
Since these choices are independent, the total number of ways to map is:

Final Calculation

We have solved the two parts of our puzzle. We have 98 ways to handle the constrained set and 4 ways to handle the free set .
According to the fundamental principle of counting, if we have two independent tasks, we multiply the number of ways to perform each. Therefore, the total number of valid functions is:
And there you have it! By breaking the problem into logical pieces, we have navigated through the constraints and arrived at the solution with clarity and confidence. The final answer is 392.

Similar Questions

JEE Main 2023 (30 January Shift 2)
LEVELJEE Main

The number of ways of selecting two numbers a and b, and such that 2 is the remainder when is divided by 23 is

(A)
186
(B)
54
(C)
108
(D)
268
JEE Main 2021 (25 February Shift 1)
LEVELJEE Main

The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5 is

JEE Main 2021 (22 July Shift 1)
LEVELBoard

If the digits are not allowed to repeat in any number formed by using the digits 0, 2, 4, 6, 8, then the number of all numbers greater than 10,000 is equal to

JEE Main 2024 (05 Apr Shift 1)
LEVELJEE Main

Let and . Then the total number of one-one maps , such that , is :

(A)
480
(B)
240
(C)
120
(D)
180
JEE Main 2023 (12 April Shift 1)
LEVELJEE Main

The number of five-digit numbers, greater than 40000 and divisible by 5, which can be formed using the digits 0, 1, 3, 5, 7 and 9 without repetition, is equal to

(A)
132
(B)
120
(C)
72
(D)
96
JEE Main 2021 (March) (18 March Shift 1)
LEVELBoard

The number of times the digit 3 will be written when listing the integers from 1 to 1000 is

JEE Main 2012
LEVELJEE Main

Let . The number of different ordered pairs (Y,Z) that can formed such that and is empty is:

(A)
(B)
3⁵
(C)
2⁵
(D)
JEE Advanced 1998
LEVELBoard

An -digit number is a positive number with exactly digits. Nine hundred distinct -digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of for which this is possible is

(A)
6
(B)
7
(C)
8
(D)
9
JEE Main 2019 (9 January)
LEVELBoard

The number of natural numbers less than 7,000 which can be formed by using the digits 0,1,3,7,9 (repetition of digits allowed) is equal to :

(A)
250
(B)
374
(C)
372
(D)
375
JEE Main 2018 (15 April Shift 1)
LEVELBoard

n-digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is

(A)
9
(B)
6
(C)
8
(D)
7