I'm a bit rusty in combinatorics, but I'll give it a shot.

Step:

- the easy ones

2 favorable base quintuples (12345 and 23456) with each having 5! permutations = 2 * 5! = 2 * 120 =

**240**- strict combinations of only 4 consecutive numbers

2 favorable quadruples (1234, 3456) with each having 4! permutations = 2 * 4! = 2 * 24 = 48

48 quadruples * 5 positions the extra dice can be * 1 number the extra dice can be = 48 * 5 =

**240**(if there are 1234, then the only number allowed is 6 because it does not continue the sequence as 5 would since that would be a quintuple which we already counted)

- strict combinations of only 3 consecutive numbers

2 favorable triples (123, 456) with each having 3! permutations = 2 * 3! = 2 * 6 = 12

12 triples * 10 positions the extra dice can be * 4 numbers the extra dice can be = 12 * 10 * 4 =

**480**2 favorable triples (234, 345) with each having 3! permutations = 2 * 3! = 2 * 6 = 12

12 triples * 10 positions the extra dice can be * 1 number the extra dice can be = 12 * 10 * 1 =

**120**(if there are 234, then the only number allowed is 6 because it does not continue the sequence as 1 or 5 would)

- combinations of 4 numbers where the 5th one is a repetition

3 favorable quadruples (1234, 2345, 3456) with each having 3! permutations = 3 * 3! = 3 * 6 = 18

18 quadruples * 10 positions the repeating dice can be * 4 numbers the extra dice can be = 18 * 10 * 4 =

**720**Now I'm not entirely sure if I am going in the right direction with this and if there is a simpler method. There are 3 more combination group that have to be counted.

- combinations of 3 numbers with one repetition

- combinations of 3 numbers with two repetitions

- combinations of 3 numbers with two repetitions of the same number

I wrote a small program to count them manually and the result should be 3480 (~45%). I have counted 1800 so far so the remaining 1680 should be in the remaining 3 groups. If I'm right that is.