Manipal Entrance Test 2015 Math Paper

There can be 2 types of numbers.
(i) Anyone of the digits 1, 2, 3, 4 repeats thrice and the remaining digits only once i.e., of the type 1, 2, 3, 4, 4, 4.
(ii) Any two of the digits 1, 2, 3, 4 repeat twice and the remaining two only once i e., of the type 1,2, 3, 3, 4, 4.
Number of numbers of the type 1 2 3 4 4 4 =

6!

3!

×4C1=480
Number of numbers of the type 1 2 3 3 4 4

6!

2!2!

×4C2=1080
So, the required number = 480 + 1080 = 1560