[Math] (SOLVED) How many strings of length 10 either start with 11 or end with 100

combinatoricsdiscrete mathematics

I'm somewhat floored as to what my professor is trying to ask here, i'm currently taking discrete mathematics and he gave the class this question for an upcoming exam. Can anybody offer an explanation?

(provided by my professor) the answer is : $ 2^8 $ + $ 2^7 $ $= 384$

i am aware my professor made a typo in the question and has also given an incorrect answer to the question he provided to the class. I will have a discussion with him next class about this question.

Best Answer

Assuming he means binary strings (containing $1$ or $0$) then there are $2^{10}$ possible strings total (you have two choices for each character). There are $2^{8}$ that start with $11$ (you have two choices for each of the remaining characters). There are $2^{7}$ that end with $100$ (ditto) and $2^{5}$ that do both (still ditto).

So there are are $2^8 - 2^5$ that start with $11$ but do not end with $100$. There are $2^7 - 2^5$ that end with $100$ but do not start with $11$. And there are $2^5$ that do both.

So together there are $2^8 + 2^7 - 2^5$ that do one or the other or both.