This is what I think:
You can either include the digit in the group, or you cannot.
There are 9 digits. You can say yes to that digit, or no to that digit. (And note there can't be repetition.)
So that's a total of 29 outcomes.
(But, if you have to have at LEAST one digit, you can't have no no no no no no no no no)
So in total 29 - 1
______________
If you were to use part (i), the trick is that you want to find 9C1 + 9C2 + 9C3 + ... + 9C1
I.e. 29 MINUS 9C0
Perfect, except there are
10 digits including 0, so that would be:
hey guys how do you do part b?
Also, in general, how do you prove that a coefficient(s) is/are the greatest in an expansion. do you just have to proves the the term in front divided by that term have a value of less than one, thus it is the largest?
In general, that method is correct! The usual idea is to take the general form of the k-th coefficient, and the general form of the (k+1)th coefficient, and take the ratio (Tk)/T_(k+1)). As soon as this is greater than one you have the k-th coefficient as greater (remember k can only be a positive integer, no decimals)
Don't quite have time to punch out a full solution for (ii), however, I think it would be the exact same thing, except instead of considering the general form of coefficients, you'd consider the general form of TERMS! So:
Do the same thing and find a ratio. There should be a value of k which makes it EQUAL to 1, and that will give your equal terms