\(\text{There are }\dbinom{4}{1}19^3 \text{ different words containing a single instance of the letter A} \\ \dbinom{4}{1} \text{ spots for the A and then a choice of 19 letters, B-T, for each of the 3 remaining spots}\\ \text{Similarly there are }\dbinom{4}{k}19^{4-k} \text{ words containing }k \text{ instances of the letter A}\\ \text{Thus we have}\\ N=\sum \limits_{k=1}^4 ~\dbinom{4}{k}19^{4-k} = 29679\)
It strikes me that perhaps a simpler way of doing this is to subtract all the words with no instances of the letter A
from the total number of words possible.
\(\text{There are }19^4 \text{ words that have no A in them}\\ \text{There are }20^4 \text{ total possible words}\\ N=20^4-19^4 = 29679\)
.
+6248 