Let the 4-digit ATM pin code be represented by the digits $$abxy$$, where all 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: $$a + b = x + y$$.

Since the greatest digit is 7, the possible digits for the pin code are $$0, 1, 2, 3, 4, 5, 6,$$ and $$7$$.

1. Let's denote the sum of the first two digits and the sum of the last two digits as $$\lambda$$. Since the largest digit is 7, we can analyze different possible values for $$\lambda$$.

We will analyze different possible values for $$\lambda$$ and the corresponding possible pairs of digits:

- $$\lambda = 7$$: The pairs of digits that have a sum of 7 are $(0,7),(7,0),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. Since 7 must be present, we consider the pairs with 7: $(0,7),(7,0)$.

When the pair $(0,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. For each of these pairs, there are 2 possible pin codes: $$07xy$$ and $$07yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 12 possible pin codes for this case.

When the pair $(7,0)$ is chosen, the other two digits can also be any one of the remaining pairs: $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. For each of these pairs, there are 2 possible pin codes: $$70xy$$ and $$70yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 12 possible pin codes for this case.

So, there are 24 possible pin codes for $$\lambda = 7$$.

JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 84 English Explanation 1

JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 84 English Explanation 1

- $$\lambda = 8$$: The pairs of digits that have a sum of 8 are $(1,7),(7,1),(2,6),(6,2),(3,5),(5,3)$. Since 7 must be present, we consider the pairs with 7: $(1,7),(7,1)$.

When the pair $(1,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(2,6),(6,2),(3,5),(5,3)$. For each of these pairs, there are 2 possible pin codes: $$17xy$$ and $$17yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.

When the pair $(7,1)$ is chosen, the other two digits can also be any one of the remaining pairs: $(2,6),(6,2),(3,5),(5,3)$. For each of these pairs, there are 2 possible pin codes: $$71xy$$ and $$71yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.

So, there are 16 possible pin codes for $$\lambda = 8$$.

JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 84 English Explanation 2

JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 84 English Explanation 2

- $$\lambda = 9$$: The pairs of digits that have a sum of 9 are $(2,7),(7,2),(3,6),(6,3),(4,5),(5,4)$. Since 7 must be present, we consider the pairs with 7: $(2,7),(7,2)$.

When the pair $(2,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(3,6),(6,3),(4,5),(5,4)$. For each of these pairs, there are 2 possible pin codes: $$27xy$$ and $$27yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.

When the pair $(7,2)$ is chosen, the other two digits can also be any one of the remaining pairs: $(3,6),(6,3),(4,5),(5,4)$. For each of these pairs, there are 2 possible pin codes: $$72xy$$ and $$72yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.

So, there are 16 possible pin codes for $$\lambda = 9$$.

JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 84 English Explanation 3

JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 84 English Explanation 3

- $$\lambda = 10$$: The pairs of digits that have a sum of 10 are $(3,7),(7,3),(4,6),(6,4)$. Since 7 must be present, we consider the pairs with 7: $(3,7),(7,3)$.

When the pair $(3,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(4,6),(6,4)$. For each of these pairs, there are 2 possible pin codes: $$37xy$$ and $$37yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.

When the pair $(7,3)$ is chosen, the other two digits can also be any one of the remaining pairs: $(4,6),(6,4)$. For each of these pairs, there are 2 possible pin codes: $$73xy$$ and $$73yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.

So, there are 8 possible pin codes for $$\lambda = 10$$.

JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 84 English Explanation 4

JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 84 English Explanation 4

- $$\lambda = 11$$: The pairs of digits that have a sum of 11 are $(4,7),(7,4),(5,6),(6,5)$. Since 7 must be present, we consider the pairs with 7: $(4,7),(7,4)$.

When the pair $(4,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(5,6),(6,5)$. For each of these pairs, there are 2 possible pin codes: $$47xy$$ and $$47yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.

When the pair $(7,4)$ is chosen, the other two digits can also be any one of the remaining pairs: $(5,6),(6,5)$. For each of these pairs, there are 2 possible pin codes: $$74xy$$ and $$74yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.

So, there are 8 possible pin codes for $$\lambda = 11$$.

JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 84 English Explanation 5

JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 84 English Explanation 5

- $$\lambda = 12$$: The only pair of digits that have a sum of 12 is $(5,7)$. Since 7 must be present, this case is not possible.

- $$\lambda = 13$$: The only pair of digits that have a sum of 13 is $(6,7)$. Since 7 must be present, this case is not possible.

- $$\lambda = 14$$: The only pair of digits that have a sum of 14 is $(7,7)$. Since all the digits must be different, this case is not possible.

Adding up all the possible pin codes for each value of $$\lambda$$, we get:

$$16 + 24 + 24 + 16 + 8 = 72$$

Therefore, the maximum number of trials necessary to obtain the correct code is 72.

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