We want to arrange digits 0 to 9 such that 1 and 2 always have 2 digits between them (we want to find the number of different combinations to do this).
There are 564,480 ways of arranging the digits 0 to 9 in such a way that there is always 2 digits between 1 and 2.
Now let's see how to get the number of combinations.
We have 10 numbers to order, so we have 10 places to put these numbers.
_ _ _ _ _ _ _ _ _ _
Let's start by assuming that the first number is the 1
1 _ _ _ _ _ _ _ _ _
Because 1 and 2 must have 2 digits between them, we put the 2 in the third place
1 _ _ 2 _ _ _ _ _ _
Now in the other 8 places we can put any of the other 8 numbers in any order, The number of combinations is just given by:
8! = 8*7*6*5*4*3*2*1 = 40,320
Now we also need to take in account the permutations of the 1 and the 2.
First, these two can permute (so the 2 goes first and then the 1) so there is a factor 2.
Also, we fixed the first one at the first position, but it could be in the second, third, fourth, fifth, sixth or seventh position, so there is also a factor 7.
Then the total number of combinations is given by:
2*7*40,320 = 564,480
This means that there are 564,480 ways of arranging the digits 0 to 9 in such a way that there is always 2 digits between 1 and 2.
If you want to learn more, you can read:
https://brainly.com/question/3732964
