Homework #10. Odd solutions in back of text.
Section 7.1: 2a) an = an-1 + an-2 + an-3, b) a5 = 13;
4a) an = an-3 + an-6 + an-10, b) a12 = 5. 6a) an = an-1 + an-2, b) an = 2an-1 + 2an-2.
10. an = an-1 + 2n-2; 12. an = an-1 + 2n - 2; 28. an,k = an,k-1 + an-1,k; 30. an,k = an-2,k-1 + an-4,k-1 + an-6,k-1.

Homework #11 Odd solutions in back of text.
Section 8.1: 10. 10! - 2 x 9! + 8!; 12a) 2x24! - 22!, b) 26! - 24! - 23! + 22!;
16a) yes - the problem here is that you get a negative answer if you use inclus-exclus formula to compute the percent who like none, b) 40%-- assume no one likes none of these recreations. With this assumption, you can treat the three-way intersection as an unknown and solve for it in the Incl-Excl formula;
24. 3 x 5! - 3 x 4! + 3!;
27.Comment: The properties to be avoided are that a family has one child in each tent. The ways for this to happen are 5! (pick a different tent for each of the, say, Bernstein children) x 10!/2!^5 (ways to distribute the other ten children from the other two families with two in each tent). Answer is: {15!/3!^5 - 3x5!x10!/2!^5 + 3x5!^3 - 5!^3}/15!/3!^5.
36. 2.
Section 8.2: 2. 6^10 - C(6,1) x 5^10 + C(6,2) x 4^10 - C(6,3) x 3^10 + C(6,4) x 2^10 - C(6,5);
8. sum for k = 0 to n: (-1)^k x C(n,k) x (2n-k)!/2!^(n-k);
14. _ sum for k = 0 to n. (-1)^k x C(n,k) x C((r - n) - 5k + n - 1, n - 1)-hint: Pick one toy of each model type and then pick remaining r-n toys with no more than 4 additional toys of any type (lilke Example 3.