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## Homework Statement

Using the common logic and known combinatorics properties solve the following problems:

a) If you have 6 letters and 6 envelopes, each having it's corresponding envelope in how many ways can you place every letter in the wrong envelope? (1 letter goes into 1 envelope)

b)You have been given a set ##A={1,2...50}##. Determine the number of subsets of A that have 5 elements, which don't include 45 and which don't include two succesive numbers.

## Homework Equations

3. The Attempt at a Solution [/B]

The first one should be relativly simple and i just wanted to check my result with you.

Lets say you have the first envelope. Number of wrong letters that you can put into it is 5 and so on for the every envelope and the result is ##5^6##. Do you agree?

The second one is what im really curious about. I thought of it the similar way i did with the upper problem.

Lets say i want to create that subset. I have 5 empty places for the elements. I can't have 49 and that leaves me with 49 possibilities. For the second place i can have 48 possibilities because i cant have 49 and the successor of the one in the first place.

That leaves me with the result ##49*48^4##. What do you think?