A person forgets his 4 -digit ATM pin code. But he remembers that in the code all the digits are different,…

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

$\lambda =8, \left(a,b\right)\to \left(1,7\right) \text{or} \left(7,1\right)$, 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

$\lambda =12, 13 \& 14$ are not possible,

So, Total numbers $=24+16+16+8+8=72$