[abdmoh!]

Hello,

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)

The question:

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?

My solution:

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..

Thanks

[starkrush]

Let A = start with 3, B = end with 4
n(AuB) = n(A) + n(B) - n(AnB).

Does that help?