Question

Consider all length-13 strings of all uppercase letters. Letters may be repeated.

Answer 1

a. Your answer is correct, 26 choices for each of 13 positions, so $\boxed{26^{13}}$ total possible strings.

b. You have the right idea, but your method only counts one type of permutation, like

C H A R I T Y _ _ _ _ _ _

but doesn't account for other arrangements like

_ _ _ C H A R I T Y _ _ _

or

_ _ _ _ _ C H A R I T Y _

Treating CHARITY as one letter, we're then considering strings of length 7 (6 open slots plus this string), which we can arrange in 7! different ways. So the total number of such strings is $\boxed{7!26^6}$.

c. This one is a bit more involved. I would go about it by counting the number of strings containing CHARITY but not HORSES, HORSES but not CHARITY, and both CHARITY and HORSES.

• CHARITY but not HORSES

As we know from part (a), there are $7!26^6$ strings containing CHARITY, but the string HORSES can be found whenever there are 6 open slots to either side of CHARITY, i.e. in strings of either form

C H A R I T Y _ _ _ _ _ _

or

_ _ _ _ _ _ C H A R I T Y

Then there are 2 strings that we want to remove from the count, giving $7!26^6-2$ such strings.

• HORSES but not CHARITY

Reasoning as we did in part (b) suggests that there are $8!26^7$ possible strings containing HORSES, and reasoning as we did in the previous case suggests only 2 of these contain CHARITY, giving a total of $8!26^7-2$ such strings.

• CHARITY and HORSES

There are 2 such strings,

C H A R I T Y H O R S E S

H O R S E S C H A R I T Y

Then by the inclusion-exclusion principle, the number of strings containing either CHARITY or HORSES is $(7!26^6-2)+(8!26^7-2)-2=7!26^6209-6$.

Finally, the number of strings containing neither CHARITY nor HORSES is complementary to the number of strings containing either of them, so the total is $26^{13}-(7!26^6209-6)=\boxed{26^{13}-7!26^6209+6}$.