A person forgets his 4 -digit ATM pin code. But he remembers that...

Let the $4 -$digit pin be $abcd$

Now given sum of first two is equal to last two, so $a + b = c + d$

And number available are $0, 1, 2, 3, 4, 5, 6, 7$

Now Let $a + b = c + d = \lambda$

Also the greatest digit that he has used is $7$

So, $a + b = c + d = \lambda \ge 7$

Now taking all cases starting from $\lambda = 7$ we get,

$\lambda = 7, \left(a, b\right) \to \left(0, 7\right)\text{or}\left(7, 0\right)$, then \left(c, d\right) \to \left(1, 6\right), \left(2, 5\right), \left(3, 4\right), \left(4, 3\right), \left(5, 2\right), \left(6, 1\right)$and vice-versa, so$2 \times 12 = 24$numbers</p><p>$\lambda = 8, \left(a, b\right) \to \left(1, 7\right)\text{or}\left(7, 1\right)$, then \left(c, d\right) \to \left(2, 6\right), \left(3, 5\right), \left(6, 2\right), \left(5, 3\right)$ and vice versa so total $2 \times 8 = 16$ numbers,

$\lambda = 9, \left(a, b\right) \to \left(2, 7\right)\text{or}\left(7, 2\right)$ then \left(c, d\right) \to $\left(3, 6\right), \left(4, 5\right), \left(5, 4\right), \left(6, 3\right)$ and vice versa, so total $2 \times 8 = 16$ numbers,

$\lambda = 10, \left(a, b\right) \to \left(3, 7\right), \left(7, 3\right), \text{so}\left(c, d\right) \to \left(6, 4\right), \left(4, 6\right) \to 8$ numbers

$\lambda = 11, \left(a, b\right) \to \left(4, 7\right), \left(7, 4\right), \text{so}\left(c, d\right) \to \left(6, 5\right), \left(5, 6\right) \to 8$ numbers

$\lambda = 12, 13 \text{ and } 14$ are not possible,

So, Total numbers $= 24 + 16 + 16 + 8 + 8 = 72$