If neither '5' nor '6' is allowed, there are strings, since there are now only 8 choices for each position. So the total number that include a '5' or '6' (or both) is .
(b) If we choose items from different ones (where we're not allowed to choose the same item more than once) and then re-arrange them in order (this is called the number of permutations of items from ), the number of ways of doing this is . We need to do this with the 26 letters, in such a way that the number of choices is at least 300,000. So we want the lowest value of for which .
Using a calculator (or other aid) the lowest value of is found by trial to be .
(c) The question is this: From a group of 13 people, a group of five is chosen. Then from the remaining 8 people a second group of five is chosen, with the remaining 3 then forming a third group. So the original group of 13 has been divided into two groups of five, plus a group of three.
Here we use to find the number of selections of items from , where the order within the selection doesn't matter. The order in which we choose our groups of 5, 5 and 3 doesn't matter either, but there is a small problem to be overcome that arises because two of the groups are the same size.
Because of this problem, I think it's easier to understand if we choose the group of 3 first. This can be done in ways.
We are left with a group of 10 that must be split into 2 groups of 5. The number of ways of choosing 5 from 10 is . But we need to realise that for every group of 5 that is chosen, another group of 5 is formed by those that are left over. So in the ways, each group will be chosen twice. So we must divide by 2 to find the number of different ways of splitting the 10 into two groups of 5. This gives ways.
So the total number of ways of forming groups of 5, 5 and 3 from the original is .