## Homework Assignment Nine Quick Solutions

```
1. 6.2.4  C(14,9)-C(9,4)

2. 6.2.18   a.C(r-1,r-8); b. C(r/2+7,r/2)

3. 6.2.24 C(89,40)-C(86,37)C(50,1)+C(83,34)C(50,2)..
That is, coefficient of x**40 in (1+x+x**2)**50

4. 6.2.30 this is: 1+(x+..+x**6)+(x+..+x**6)**2..

5. 6.3.14 Let a+b+c be a partition of N. Then
(a+b)+(a+c)+(b+c) is a partition of 2N of the correct
form. Furthermore, given a parition of 2N of the correct
form, it’s not hard to recover a unique a-b-c that make
up the N partition.

6. 6.3.22 Let S(n,k) stand for the notation in the book.
If one new object is added to the mix, then it can join
any of the existing k partitions, or it can stand alone,
in which case S(n,k-1) partitions of the other n objects
are possible.

7. 6.4.10 a. Expand (e**x-x**2/2)**3. This yields for
coefficient of x**r: 3**r-(3*2**(r-3)*(r(r-1)))+(3r(r-
1)(r-2)(r-3))/4) – (6!)/8, where the last term
participates only when there are exactly 6 digits in
number. Inclusion/Exclusion solution provides more
insight here. B. expand ((1/2(e**x+e**-x)-1)**2)e**x

8. 6.5.2e: 4!C(n+1,5)

9. 7.1.2 An = An-1 + An-2 + An-3

```

-- RobbieMoll - 05 Dec 2006

Topic revision: r1 - 2006-12-05 - RobbieMoll

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