JEE Main 2025 (January)

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Animated Solution for Mathematics - Permutations and Combinations: The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is

Q: Wait, why do we subtract 1 from $a$ at the start?

Great question! The 'Stars and Bars' formula $\binom{n+r-1}{r-1}$ works specifically for non-negative integer solutions (where variables can be 0). Since our hundreds digit $a$ must be at least 1, we define $a' = a-1$ so that $a' \ge 0$. This simple shift allows us to use the formula correctly.

Q: Why only check $a=1$ and $a=2$ for the range constraint?

We need numbers $> 212$. For $a=1$, the sum $1+b+c=15$ means $b+c=14$. Some of these combinations (like $159$) are $\le 212$. For $a=2$, the smallest possible number is $249$ (when $b=4, c=9$), which is already greater than $212$. Thus, we only need to filter out the few cases starting with 1.

Enter Numerical Value:

Visualized Solution

Problem Analysis and Constraints

Find $n \in (212, 999)$ such that sum of digits is $15$ .
Let the number be $ab c = 100 a + 10 b + c$ .
Condition: $a + b + c = 15$ .
Constraints: $a \in {1, 2, \dots, 9}$ , $b, c \in {0, 1, \dots, 9}$ .

Variable Transformation for $a \geq 1$

To use Stars and Bars, variables must be $\geq 0$ .
Let $a^{'} = a - 1$ , where $a^{'} \geq 0$ .
Substituting $a = a^{'} + 1$ into $a + b + c = 15$ :
$(a^{'} + 1) + b + c = 15 \Rightarrow a^{'} + b + c = 14$ .

Applying Stars and Bars Formula

Formula for non-negative integer solutions to $x_{1} + x_{2} + \dots + x_{r} = n$ :
Number of solutions = $^{n + r - 1} C_{r - 1}$ .
Here, $n = 14$ and $r = 3$ .
Total solutions = $^{14 + 3 - 1} C_{3 - 1} =^{16} C_{2}$ .

Calculating Total Unconstrained Solutions

$^{16} C_{2} = \frac{16 \times 15}{2 \times 1} = 120$ .
These are all solutions where $a^{'}, b, c \geq 0$ .
Constraint Check: Since $a, b, c$ are digits, $a, b, c \leq 9$ .
We must subtract cases where $a > 9$ , $b > 9$ , or $c > 9$ .

Handling Constraint Violation for $a$

Case 1: $a > 9 \Rightarrow a \geq 10 \Rightarrow a^{'} \geq 9$ .
Let $a^{''} = a^{'} - 9 \geq 0$ .
Equation: $(a^{''} + 9) + b + c = 14 \Rightarrow a^{''} + b + c = 5$ .
Solutions = $^{5 + 3 - 1} C_{3 - 1} =^{7} C_{2} = 21$ .

Handling Constraint Violation for $b$

Case 2: $b > 9 \Rightarrow b \geq 10$ .
Let $b^{'} = b - 10 \geq 0$ .
Equation: $a^{'} + (b^{'} + 10) + c = 14 \Rightarrow a^{'} + b^{'} + c = 4$ .
Solutions = $^{4 + 3 - 1} C_{3 - 1} =^{6} C_{2} = 15$ .

Handling Constraint Violation for $c$

Case 3: $c > 9 \Rightarrow c \geq 10$ .
Let $c^{'} = c - 10 \geq 0$ .
Equation: $a^{'} + b + (c^{'} + 10) = 14 \Rightarrow a^{'} + b + c^{'} = 4$ .
Solutions = $^{4 + 3 - 1} C_{3 - 1} =^{6} C_{2} = 15$ .

Net Valid 3-digit Numbers

Total 3-digit numbers with sum 15 = Total - (Violations).
Valid Count = $120 - (21 + 15 + 15)$ .
Valid Count = $120 - 51 = 69$ .

Identifying Numbers $\leq 212$ with $a = 1$

We need numbers $> 212$ . Let's find numbers $\leq 212$ with sum 15.
If $a = 1$ , then $1 + b + c = 15 \Rightarrow b + c = 14$ .
Possible $(b, c)$ pairs: $(5, 9), (6, 8), (7, 7), (8, 6), (9, 5)$ .
These are 5 numbers: $159, 168, 177, 186, 195$ .

Checking Range for $a = 2$

If $a = 2$ , then $2 + b + c = 15 \Rightarrow b + c = 13$ .
Smallest number is $249$ (when $b = 4, c = 9$ ).
Since $249 > 212$ , all numbers with $a \geq 2$ are within the range.
Only the 5 numbers from the $a = 1$ case need to be excluded.

Final Calculation of Valid Numbers

Total valid numbers = (Total 3-digit numbers with sum 15) - (Numbers $\leq 212$ ).
Total = $69 - 5 = 64$ .
Final Answer: $64$ .

00:00 / 00:00

The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

We represent the three-digit number as

ab c

, which corresponds to the value

100 a + 10 b + c

. The constraints are defined by the sum of the digits:

a + b + c = 15

Here, the digit

a

must satisfy

a \in {2, 3, \dots, 9}

to ensure the number is at least 212, while

b, c \in {0, 1, \dots, 9}

. We will first calculate the total solutions for

a \in {1, \dots, 9}

and then subtract the invalid cases.

The Stars and Bars Transformation

To apply the 'Stars and Bars' method, we require non-negative variables. We define

a^{'} = a - 1

, such that

a^{'} \geq 0

. Substituting this into the sum:

(a^{'} + 1) + b + c = 15 ⟹ a^{'} + b + c = 14

The number of non-negative integer solutions is given by the formula

(r - 1 n + r - 1)

, where

n = 14

and

r = 3

(3 - 1 14 + 3 - 1) = (2 16) = 120

The Constraint Trap

The 120 solutions include cases where digits exceed 9. We use the Principle of Inclusion-Exclusion to remove these invalid cases.

a \geq 10

, let

a^{''} = a^{'} - 9

. The equation becomes

(a^{''} + 9) + b + c = 14

, or

a^{''} + b + c = 5

. The number of invalid solutions is:

(3 - 1 5 + 3 - 1) = (2 7) = 21

b \geq 10

, let

b^{'} = b - 10

. The equation becomes

a^{'} + (b^{'} + 10) + c = 14

, or

a^{'} + b^{'} + c = 4

. The number of invalid solutions is:

(3 - 1 4 + 3 - 1) = (2 6) = 15

By symmetry, if

c \geq 10

, there are also 15 invalid solutions. Subtracting these from the total:

120 - (21 + 15 + 15) = 69

The Final Filter

We have 69 valid three-digit numbers where

a \geq 1

. However, we must exclude numbers less than or equal to 212.

a = 1

, then

b + c = 14

. The possible pairs

(b, c)

are

(5, 9), (6, 8), (7, 7), (8, 6), (9, 5)

, totaling 5 numbers: 159, 168, 177, 186, and 195.

a = 2

, the smallest number with a digit sum of 15 is 249, which is already greater than 212. Thus, we only subtract the 5 numbers identified when

a = 1

The final count is:

69 - 5 = 64

The total number of valid integers is 64.

Common Traps & FAQs

Wait, why do we subtract 1 from

a

at the start?

Great question! The 'Stars and Bars' formula

(r - 1 n + r - 1)

works specifically for non-negative integer solutions (where variables can be 0). Since our hundreds digit

a

must be at least 1, we define

a^{'} = a - 1

so that

a^{'} \geq 0

. This simple shift allows us to use the formula correctly.

Why do we subtract 51 from the total?

That 51 represents the 'invalid' cases where our digits violate the rule of being single digits (i.e.,

d > 9

). We used the Principle of Inclusion-Exclusion: we calculated cases where

a \geq 10

b \geq 10

, and

c \geq 10

separately and subtracted them from the total to ensure every digit remains between 0 and 9.

Why only check

a = 1

and

a = 2

for the range constraint?

We need numbers

> 212

. For

a = 1

, the sum

1 + b + c = 15

means

b + c = 14

. Some of these combinations (like

159

) are

\leq 212

. For

a = 2

, the smallest possible number is

249

(when

b = 4, c = 9

), which is already greater than

212

. Thus, we only need to filter out the few cases starting with 1.

Animated Solution for Mathematics - Permutations and Combinations: The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is

Enter Numerical Value:

Visualized Solution

Problem Analysis and Constraints

Variable Transformation for $a \geq 1$

Applying Stars and Bars Formula

Calculating Total Unconstrained Solutions

Handling Constraint Violation for $a$

Handling Constraint Violation for $b$

Handling Constraint Violation for $c$

Net Valid 3-digit Numbers

Identifying Numbers $\leq 212$ with $a = 1$

Checking Range for $a = 2$

Final Calculation of Valid Numbers

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Stars and Bars Transformation

The Constraint Trap

The Final Filter

Similar Questions

The total number of 3 -digit numbers, whose sum of digits is 10, is

The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is _________

The number of triplets $(x, y, z)$ where $x, y, z$ are distinct non negative integers satisfying $x + y + z = 15$ , is

The total number of three-digit numbers, with one digit repeated exactly two times, is ______.

The number of seven digit positive integers formed using the digits 1, 2, 3 and 4 only and sum of the digits equal to 12 is _______.

Let S denote the set of 4-digit numbers $ab c d$ such that $a > b > c > d$ and P denote the set of 5-digit numbers having product of its digits equal to 20. Then $n (S) + n (P)$ is equal to ......... .

The number of numbers between 2,000 and 5,000 that can be formed with the digits 0,1,2,3,4 (repetition of digits is not allowed) and are multiple of 3 is :-

If all the six digit numbers $x_{1} x_{2} x_{3} x_{4} x_{5} x_{6}$ with $0 < x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6}$ are arranged in the increasing order, then the sum of the digits in the $7 2^{t h}$ number is ____.

Let $S = {1, 2, 3, 5, 7, 10, 11}$ . The number of non-empty subsets of $S$ that have the sum of all elements a multiple of 3, is ____.

The total number of functions, $f : {1, 2, 3, 4} \to {1, 2, 3, 4, 5, 6}$ such that $f (1) + f (2) = f (3)$ , is equal to:

Animated Solution for Mathematics - Permutations and Combinations: The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is

Enter Numerical Value:

Visualized Solution

Problem Analysis and Constraints

Variable Transformation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a≥1

Applying Stars and Bars Formula

Calculating Total Unconstrained Solutions

Handling Constraint Violation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Handling Constraint Violation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Handling Constraint Violation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Net Valid 3-digit Numbers

Identifying Numbers .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ≤212 with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=1

Checking Range for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=2

Final Calculation of Valid Numbers

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Stars and Bars Transformation

The Constraint Trap

The Final Filter

Similar Questions

The total number of 3 -digit numbers, whose sum of digits is 10, is

The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is _________

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of triplets (x,y,z) where x,y,z are distinct non negative integers satisfying x+y+z=15, is

The total number of three-digit numbers, with one digit repeated exactly two times, is ______.

The number of seven digit positive integers formed using the digits 1, 2, 3 and 4 only and sum of the digits equal to 12 is _______.

The number of numbers between 2,000 and 5,000 that can be formed with the digits 0,1,2,3,4 (repetition of digits is not allowed) and are multiple of 3 is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={1,2,3,5,7,10,11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is ____.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The total number of functions, f:{1,2,3,4}→{1,2,3,4,5,6} such that f(1)+f(2)=f(3), is equal to:

Variable Transformation for $a \geq 1$

Handling Constraint Violation for $a$

Handling Constraint Violation for $b$

Handling Constraint Violation for $c$

Identifying Numbers $\leq 212$ with $a = 1$

Checking Range for $a = 2$

The number of triplets $(x, y, z)$ where $x, y, z$ are distinct non negative integers satisfying $x + y + z = 15$ , is

Let $S = {1, 2, 3, 5, 7, 10, 11}$ . The number of non-empty subsets of $S$ that have the sum of all elements a multiple of 3, is ____.

The total number of functions, $f : {1, 2, 3, 4} \to {1, 2, 3, 4, 5, 6}$ such that $f (1) + f (2) = f (3)$ , is equal to: