We are going to count the number of six-letter strings that contain exactly one vowel. The strategy is:
First: The number of ways a single vowel can be inserted into a six letter string is $C(6,1) = 6$ -- i.e. we can put the vowel in one of six positions.
Second: Once the position is determined, there are five vowels from which to choose, so $C(5,1) = 5$ ways to insert a vowel with a determined position.
Third: The remaining five positions are a free choice of 21 letters, so there are $21^5$ ways to do this.
In all, then, the total number of strings being described is $6 \times 5 \times 21^5$:
6*5*21^5
122523030
We are now going to count the number of six-letter strings that have exactly two vowels. We will use the same strategy as before:
This time the number of positions we can choose is 2 out of 6, or $C(6,2)$
The number of ways to fill those two positions is $5^2$, since they are a free choice of two vowels and we can repeat the vowel if we want.
Finally, as above, the remaining four positions have $21^4$ possible options since it's a free choice of four letters out of 21.
So the total number of strings this time is $C(6,2) \times 5^2 \times 21^4$:
binomial(6,2) * 5^2 * 21^4
72930375
The number of strings with at least one vowel is the complement of the number of strings with no vowels. There are $26^6$ strings that we can possibly make, and $21^6$ of those have no vowels. Therefore the number of strings we want is $26^6 - 21^6$:
26^6 - 21^6
223149655
The number of strings with at least two vowels is $26^6$ (the total number of strings possible) minus the number of strings with no vowels, minus the number of strings with exactly one vowel. There are $21^6$ strings with no vowels. Above, we determined there were $6 \times 5 \times 21^5$ strings with exactly one vowel. So the number of strings we want is $26^6 - 21^6 - 6 \times 5 \times 21^5$:
26^6 - 21^6 - 6*5*21^5
100626625
This is simply the number of ways to choose 5 objects from a set of 52, which is $C(52,5)$:
binomial(52,5)
2598960
Question: Why isn't it $52 \times 51 \times 50 \times 49 \times 48$? It's because this product counts the number of ways that a selection of 5 objects from a group of 52 can be rearranged. It's as if you had a race with 52 competitors, and you wanted to know the number of ways to award first through fifth place. The same group of people arranged in a different order would be counted differently. But this is not the case with a hand of cards -- the order of the arrangement doesn't matter. For example a hand consisting of 4 of spades, 3 of clubs, 3 of diamonds, 10 of hearts, and ace of spades is the same had as 4 of spades, ace of spades, 3 of diamonds, 10 of hearts, and 3 of clubs. The product listed here would be exactly $5!$ times too large because it takes ordering into account when it shouldn't.
According to the Binomial Theorem, this coefficient would be $C(201, 100)$:
#3
binomial(201,100)
180200509365116430834121184084894227116588341829287927773320