Dr. Math's four-digit house number ABCD contains no zeroes and can be split into two different two-digit primes ``AB'' and ``CD'' where the digits A, B, C and D are not necessarily distinct. If each of the two-digit primes is less than 40, how many such house numbers are possible?
The two digit primes < 40 are
11, 13, 17, 19, 23, 29, 31, 37
Any two of these eight can be selected = C(8,2) = 28
And each of these can be arranged in two ways
So...the total number possible house numbers is 2 * 28 = 56