problem 3

(a)

Fred has

10

choices for Monday,

9

choices for Tuesday,

8

choices for Wednesday,

7

choices for Thursday and

6

choices for Friday.

10 \times 9 \times 8 \times 7 \times 6

(b)

For the first restaurant, Fred has

10

choices. For all subsequent days, Fred has

9

choices, since the only restriction is that he doesn’t want to eat at the restaurant he ate at the previous day.

10 \times 9^{4}