Can someone please explain to me how this concept works and how it applies to this question:
n(AuB) = n(A) + n(B) - n(AuB)
How many three digit codes can be made using the digits 0,1,2,3,4,5 if there are no restrictions on using a digit more than once?
How many of these codes:
f)start with a 3 or end with a 4?
1x6x6 + 6x6x1 - (1x6x1) ---> answer is correct but i dont understand the n(AuB) part so please can you help me, i understand that the events are mutually exclusive so yeah..
Let A = start with 3, B = end with 4
n(AuB) = n(A) + n(B) - n(AnB).
Does that help?
Thanks for posting! You just need to create an account in order to submit the post
Already a member?
Oops, something wasn't right
please check the following:
Not got an account?
Sign up now
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.