How do I know I haven't missed any cases for $\lambda$?

Great question! We checked $\lambda$ starting from the minimum ($7+0=7$) up to the point where we can no longer form two distinct pairs. Once $\lambda$ reaches $12$, the only pair containing $7$ is $\{5, 7\}$. Since we need another pair to satisfy the sum condition, and no other distinct digits sum to $12$ without using $7$ or repeating digits, we know we have exhausted all possibilities.

JEE Main 2023 (15 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Permutations and Combinations: A person forgets his 4-digit ATM pin code. But he remembers that in the code all the digits are different, the greatest digit is 7 and the sum of the first two digits is equal to the sum of the last two digits. Then the maximum number of trials necessary to obtain the correct code is ________.

Q: Why do we multiply by 2 when we have two pairs?

Think of the PIN as two slots: the first two digits $(a, b)$ and the last two digits $(c, d)$. If you have two pairs, say $P_1$ and $P_2$, you can place $P_1$ in the first slot and $P_2$ in the second, OR $P_2$ in the first and $P_1$ in the second. That is the factor of 2. Then, within each pair, the two digits can swap places, giving another $2 \times 2 = 4$ arrangements.

Q: Does the order of the digits within the pairs matter?

Yes, absolutely! The PIN $1234$ is different from $2134$. When we calculate the arrangements, we account for the internal permutation of each pair ($2! = 2$) and the permutation of the pairs themselves ($2! = 2$). This is why we multiply by $2 \times 2 \times 2 = 8$ for each valid combination of two pairs.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: Visualize the PIN as a set of four distinct digits

a, b, c, d

. Since the greatest digit is

7

, the digit

7

must appear exactly once. The constraint

a + b = c + d = λ

implies that the set of four digits must be partitionable into two pairs with the same sum.

Hint 2

Hint 2: You need to systematically iterate through possible values of the sum

λ

. Since

7

is the largest digit, what is the minimum possible sum

λ

can take? What is the maximum? Remember, all digits must be distinct, so you cannot use the same digit twice in a pair or across pairs.

Hint 3

Hint 3: For a fixed sum

λ

, identify all possible pairs of distinct digits from the set

{0, 1, 2, 3, 4, 5, 6, 7}

that sum to

λ

. If you have

n

such pairs, you need to choose

2

pairs to form the PIN. The number of ways to arrange these is

2 \times (2!) \times (2!) = 8

for each combination of two pairs.

Visualized Solution

Visualizing the $4$ -Digit PIN

Let the $4$ -digit PIN be $ab c d$ .
Constraints:
1. All digits are distinct: $a \neq = b \neq = c \neq = d$ .

Constraint: Greatest Digit is $7$

The greatest digit is $7$ .
This means the set of available digits is ${0, 1, 2, 3, 4, 5, 6, 7}$ .
The digit $7$ must be present in the PIN.

Constraint: Sum of Pairs

The sum of the first two digits equals the sum of the last two digits.
$a + b = c + d$

Defining the Sum $λ$

Let $a + b = c + d = λ$ .
Since $7$ must be one of the digits, the minimum possible sum is $7 + 0 = 7$ .
Therefore, $λ \geq 7$ .

Case 1: $λ = 7$

Case 1: $λ = 7$ .
Possible pairs from ${0, 1, 2, 3, 4, 5, 6, 7}$ that sum to $7$ :
${0, 7}, {1, 6}, {2, 5}, {3, 4}$ .
The pair ${0, 7}$ is mandatory to include the digit $7$ .

Case 1 Calculation

If ${a, b} = {0, 7}$ , they can be arranged in $2$ ways.
${c, d}$ can be chosen from the remaining $3$ pairs ( $3$ ways) and arranged in $2$ ways.
Ways = $2 \times 3 \times 2 = 12$ .
If ${c, d} = {0, 7}$ , similarly we get $12$ ways.
Total for $λ = 7$ is $12 + 12 = 24$ .

Case 2: $λ = 8$

Case 2: $λ = 8$ .
Possible pairs: ${1, 7}, {2, 6}, {3, 5}$ .
Note: ${4, 4}$ is excluded because digits must be distinct.
Mandatory pair: ${1, 7}$ .

Case 2 Calculation

Mandatory pair ${1, 7}$ can be ${a, b}$ or ${c, d}$ ( $2$ positions).
It can be arranged in $2$ ways.
The other pair is chosen from the remaining $2$ pairs ( $2$ ways) and arranged in $2$ ways.
Total ways = $2 \times (2 \times 2 \times 2) = 16$ .

Case 3: $λ = 9$

Case 3: $λ = 9$ .
Possible pairs: ${2, 7}, {3, 6}, {4, 5}$ .
Mandatory pair: ${2, 7}$ .
Calculation is identical to $λ = 8$ .
Total ways = $16$ .

Case 4: $λ = 10$

Case 4: $λ = 10$ .
Possible pairs: ${3, 7}, {4, 6}$ .
Note: ${5, 5}$ is excluded.
We only have $2$ valid pairs, so both must be used.

Case 4 Calculation

The pairs can be placed in $2$ ways (either ${a, b}$ or ${c, d}$ ).
Digits within each pair can be arranged in $2$ ways.
Total ways = $2 \times 2 \times 2 = 8$ .

Case 5: $λ = 11$

Case 5: $λ = 11$ .
Possible pairs: ${4, 7}, {5, 6}$ .
Both pairs must be used.
Calculation is identical to $λ = 10$ .
Total ways = $8$ .

Why no more cases?

Checking $λ \geq 12$ :
For $λ = 12$ , the only valid pair is ${5, 7}$ . We lack a second pair.
For $λ = 13$ , only ${6, 7}$ is possible.
Thus, no cases exist for $λ \geq 12$ .

Final Calculation

Total number of trials = Sum of all cases.
Total = $24 + 16 + 16 + 8 + 8$ .
Total = $72$ .
Final Answer: 72

00:00 / 00:00

The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

We are tasked with finding the number of

4

-digit sequences

ab c d

such that all digits are distinct, the largest digit is

7

, and the sum of the first two digits equals the sum of the last two. Let the set of available digits be

S = {0, 1, 2, 3, 4, 5, 6, 7}

We define the "magic sum"

λ

such that:

a + b = c + d = λ

Since the largest digit is

7

, the smallest possible sum is

7 + 0 = 7

. We must systematically evaluate each possible value of

λ

while ensuring

7

is included in one of the pairs.

Systematic Search for $λ$

For

λ = 7

, the available pairs are

{0, 7}, {1, 6}, {2, 5}, {3, 4}

. To satisfy the condition, we must select the pair

{0, 7}

and one of the remaining

3

pairs.

The number of ways to arrange these is:

3 (pair choices) \times 2 (positioning of pairs) \times 2 \times 2 (internal permutations) = 24 ways.

For

λ = 8

, the pairs are

{1, 7}, {2, 6}, {3, 5}

. We must select

{1, 7}

and one of the

2

remaining pairs.

The number of ways is:

2 \times 2 \times 4 = 16 ways.

For

λ = 9

, the pairs are

{2, 7}, {3, 6}, {4, 5}

. We must select

{2, 7}

and one of the

2

remaining pairs.

The number of ways is:

2 \times 2 \times 4 = 16 ways.

Tightening Constraints

For

λ = 10

, the pairs are

{3, 7}

and

{4, 6}

. We must use both pairs to form the PIN.

The number of ways is:

1 \times 2 \times 4 = 8 ways.

For

λ = 11

, the pairs are

{4, 7}

and

{5, 6}

. Again, we must use both pairs.

The number of ways is:

1 \times 2 \times 4 = 8 ways.

For

λ = 12

, the only pair containing

7

{5, 7}

. However, there is no remaining pair that sums to

12

using the available digits

{0, 1, 2, 3, 4, 6}

, so no further solutions exist.

Final Calculation

To find the total number of valid PINs, we sum the results from each case:

Total = 24 + 16 + 16 + 8 + 8

Total = 72

The total number of possible PINs is $72$ .

Common Traps & FAQs

Wait, why can't we use the digit 8 or 9? The problem says the greatest digit is 7, but does that mean we can't use 8 or 9?

Exactly! The constraint 'the greatest digit is 7' acts as a boundary. It implies that the set of available digits is restricted to

{0, 1, 2, 3, 4, 5, 6, 7}

. If you used an 8 or a 9, the 'greatest digit' would no longer be 7, violating the premise of the problem.

Why do we multiply by 2 when we have two pairs?

Think of the PIN as two slots: the first two digits

(a, b)

and the last two digits

(c, d)

. If you have two pairs, say

P_{1}

and

P_{2}

, you can place

P_{1}

in the first slot and

P_{2}

in the second, OR

P_{2}

in the first and

P_{1}

in the second. That is the factor of 2. Then, within each pair, the two digits can swap places, giving another

2 \times 2 = 4

arrangements.

How do I know I haven't missed any cases for

λ

Great question! We checked

λ

starting from the minimum (

7 + 0 = 7

) up to the point where we can no longer form two distinct pairs. Once

λ

reaches

12

, the only pair containing

7

{5, 7}

. Since we need another pair to satisfy the sum condition, and no other distinct digits sum to

12

without using

7

or repeating digits, we know we have exhausted all possibilities.

Does the order of the digits within the pairs matter?

Yes, absolutely! The PIN

1234

is different from

2134

. When we calculate the arrangements, we account for the internal permutation of each pair (

2! = 2

) and the permutation of the pairs themselves (

2! = 2

). This is why we multiply by

2 \times 2 \times 2 = 8

for each valid combination of two pairs.

Is it possible for the sum of the first two digits to be equal to the sum of the last two digits if the digits are not distinct?

The problem explicitly states that 'all the digits are different.' This is a crucial constraint. If digits could repeat, the number of trials would be significantly higher. Always re-read the problem to ensure you haven't missed a 'distinct' or 'different' keyword!

Enter Numerical Value:

Visualized Solution

Visualizing the $4$ -Digit PIN

Constraint: Greatest Digit is $7$

Constraint: Sum of Pairs

Defining the Sum $λ$

Case 1: $λ = 7$

Case 1 Calculation

Case 2: $λ = 8$

Case 2 Calculation

Case 3: $λ = 9$

Case 4: $λ = 10$

Case 4 Calculation

Case 5: $λ = 11$

Why no more cases?

Final Calculation

The Sigma Insight: Combinations and Selection

Analyzing the Setup

Systematic Search for $λ$

Tightening Constraints

Final Calculation

Similar Questions

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 ways of getting a sum 16 on throwing a dice four times is______

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

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

A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at most three of them are red 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 ......... .

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

Enter Numerical Value:

Visualized Solution

Visualizing the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4-Digit PIN

Constraint: Greatest Digit is .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 7

Constraint: Sum of Pairs

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

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=7

Case 1 Calculation

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=8

Case 2 Calculation

Case 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=9

Case 4: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=10

Case 4 Calculation

Case 5: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=11

Why no more cases?

Final Calculation

The Sigma Insight: Combinations and Selection

Analyzing the Setup

Systematic Search for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Tightening Constraints

Final Calculation

Similar Questions

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 ways of getting a sum 16 on throwing a dice four times is______

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

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

A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at most three of them are red is __________.

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

Visualizing the $4$ -Digit PIN

Constraint: Greatest Digit is $7$

Defining the Sum $λ$

Case 1: $λ = 7$

Case 2: $λ = 8$

Case 3: $λ = 9$

Case 4: $λ = 10$

Case 5: $λ = 11$

Systematic Search for $λ$