A few simple exercies I need someone to explain to me how to do them.

**RoseEsque** · June 22nd 2012, 12:49 AM

Hello, I want to learn DM by myself, and asked a university prof. (different faculty) to give me a few exercises to solve.

Can You point me to subjects those are from and what should I learn as well as explain how those should be solved (with the solving itself done if possible)?

Use generating functions to find a_n if :a_n = a_{n - 1} + 3n, for n >= 1

Find the number of ways to select 8 balls from the set of 5 identical red balls, 3 identical
yellow balls and 7 identical green balls.

Use the extended version of the Euclid's algorithm to compute EXT-EUCLID(24,42)

How many ways are there to distribute 18 balls among 6 different persons if
a) each ball is different and each person should get 3 balls
b) all balls are identical ?

How many permutations of the letters a, a, a, b, b, b, c, c, c, d, d, d, are there with no three
consecutive letters the same?

Show that P(n, k) = P(n - 1; k - 1) + P(n - k; k) for each 1 < k < n.

**Prove It** · June 22nd 2012, 12:54 AM

For the second one, you want to know how many combinations of 8 balls you can get from the 15. This is evaluated using

$\displaystyle \begin{align*} {15\choose{8}} = \frac{15!}{8!(15 - 8)!} \end{align*}$

**RoseEsque** · June 22nd 2012, 01:14 AM

Originally Posted by Prove It

For the second one, you want to know how many combinations of 8 balls you can get from the 15. This is evaluated using

$\displaystyle \begin{align*} {15\choose{8}} = \frac{15!}{8!(15 - 8)!} \end{align*}$

Thats it? There is no trick to it? When I saw that exercise I though there's gotta be a trick to it. Don't the colours make it harder?

**Prove It** · June 22nd 2012, 01:59 AM

Originally Posted by RoseEsque

Thats it? There is no trick to it? When I saw that exercise I though there's gotta be a trick to it. Don't the colours make it harder?

No because you aren't asked for a particular combination of colours, so essentially you're just choosing 8 balls from the 15 you are given.

**BobP** · June 22nd 2012, 05:07 AM

For the first problem, and without any specialist knowledge of methods, the following approach seems to work.
The first term in the sequence $a_{0}$ is unknown so call it $a_{0}$ .
Then, by repeated substitution,
$a_{1}=a_{0}+3$
$a_{2}=a_{0}+9,$
$a_{3}=a_{0}+18,$
$a_{4}=a_{0}+30,$
$a_{5}=a_{0}+45,$
and so on.
It appears then as if we need to find some way of generating the number sequence $3,9,18,30,45$ etc.
A difference table shows that the sequence comes from a quadratic so assume that $f(n)=a*n^{2}+b*n+c,$
substitute for three values of $n$ and solve for $a,b,c.$
(Ans: $a_{n}=a_{0}+3(n^{2}+n)/2.$ )

**Soroban** · June 22nd 2012, 08:13 AM

Hello, RoseEsque!

How many ways are there to distribute 18 balls among 6 different persons if

a) each ball is different and each person gets 3 balls?

${18\choose3,3,3,3,3,3} \:=\:\frac{18!}{(3!)^6}$

b) all balls are identical and each person gets 3 balls?

The balls are identical.
Give any three balls to each of the six people.
. . There is one way.

**

**awkward** · June 24th 2012, 11:17 AM

Originally Posted by Prove It

For the second one, you want to know how many combinations of 8 balls you can get from the 15. This is evaluated using

$\displaystyle \begin{align*} {15\choose{8}} = \frac{15!}{8!(15 - 8)!} \end{align*}$

Hi Prove It,

Are you sure this is what the problem is asking?
It seems to me that it is asking how many triples (r, y, g) there are that satisfy
$r+y+g = 8$
where
$0 \le r \le 5$
$0 \le y \le 3$
$0 \le g \le 7$

**Plato** · June 24th 2012, 11:54 AM

Originally Posted by RoseEsque

Hello, I want to learn DM by myself, and asked a university prof. (different faculty) to give me a few exercises to solve. Can You point me to subjects those are from and what should I learn as well as explain how those should be solved (with the solving itself done if possible)?
Find the number of ways to select 8 balls from the set of 5 identical red balls, 3 identical
yellow balls and 7 identical green balls.

Originally Posted by awkward

It seems to me that it is asking how many triples (r, y, g) there are that satisfy
$r+y+g = 8$
where
$0 \le r \le 5$
$0 \le y \le 3$
$0 \le g \le 7$

@awkward, You are exactly right. That is what the question means.
The answer is the coefficient of $x^{8}$ in this expansion.
The clue is that right above the question is about generating functions.

**awkward** · June 24th 2012, 01:45 PM

Originally Posted by RoseEsque

Hello, I want to learn DM by myself, and asked a university prof. (different faculty) to give me a few exercises to solve.

Can You point me to subjects those are from and what should I learn as well as explain how those should be solved (with the solving itself done if possible)?

Use generating functions to find a_n if :a_n = a_{n - 1} + 3n, for n >= 1

[snip]

BobP has already given you a solution to the first part, and I think he
is right that it's easier witout generating functions in this case, but since the
problem statement says to use generating functions, here is a go at
it. I'm going to assume you know something about infinite series.

The standard approach is to let
$f(x) = \sum_{n=0}^{\infty} a_n x^n$
Using the recurrence given, we will find a closed form for f(x), and
then use that to find an expression for $a_n$ .

First, we would like to find the generating function for
$3n$ , i.e., a closed form for
$g(x) = \sum_{n=1}^{\infty} 3n x^n$ . A standard trick here
is to start with
$(1-x)^{-1} = \sum_{n=0}^{\infty} x^n$ .
Differentiate w.r.t. x, then multiply by $3x$ . The result is
$g(x) = 3x (1-x)^{-2}$ .

Now we apply the recurrence relation $a_n = a_{n-1} + 3n$ for
$n \ge 1$ . If we interpret this in terms of the generating
functions above, it becomes
$f(x) = a_0 + x f(x) + 3 x (1-x)^{-2}$
(Write out a few terms of the infinite series for f and g if you have
trouble seeing this.)
Solving for f(x),

$f(x) = a_0 (1-x)^{-1} + 3 x (1-x)^{-3}$

This is the closed form for f(x) we were looking for. The next step
is to use this to find a closed form for $a_n$ . To do this,
we will apply the binomial theorem for negative exponents (twice) on
the right-hand side of the equation. The result is

$f(x) = a_0 \sum_{i=0}^{\infty} x^i + 3x \sum_{j=0}^{\infty} \binom{j+2}{j} x^j$

It is now possible to "read off" the coefficient of $x^n$ on
the right-hand side of the equation: it is

$a_n = a_0 + 3 \binom{n+1}{n} = a_0 + \frac{3}{2} n (n+1)$ ,

which is the same result found by BobP.

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