1. ## Combinatorics

I'm trying to prepare for a course in discreet mathematics I'll have the next semester so I got a bunch of questions from a friend to help me. So far I've gone trough logic and sets. With this problems presented my biggest problem is that I dont quite understand the questions or how to go about them.

My first problem is with proofs, as I dont really understand what counts as a proof.

Well, on to the problems. In one problem I got a 16 letter long word and want to find out how many strings of length 16 I can create out of this word. Some letters apear multiple times in the word. The word I have is schoolmistresses.

In another problem I got the equation x+y+5z+w=15 and I got to find out how many solution this equation got. I'm not sure how to do this in itself and especially not with the conditions : x; y; z and w are non-negative integers and we also have that x is positive?

What I dont understand in the condition is that first we say that the variables must be non negative integers , for example 1,2,3,4,5... and then we have that x must be positive. Isn't that redundant or is there something I'm missing.

The third problem is where I must prove something.

We assume the following statement is true: there are inifnitely many positive integers Mk
such that all the binomial coefﬁcients C(Mk; j); 1 <= j <=Mk - 1 are even integers

Prove that there are also inifnitely many positive integers nk such that all the binomial coefﬁcients C(Nk; j); 0<=j <= Nk are odd integers.

In this part I understand that I should first start with understanding what a binominal coeffient is. Also if I understood the first sentence correct, the expression C(Mk,j) is always true for 1<=j<=Mk-1. Do I read this correct.

2. ## Re: Combinatorics

Originally Posted by dipsy34
Well, on to the problems. In one problem I got a 16 letter long word and want to find out how many strings of length 16 I can create out of this word. Some letters apear multiple times in the word. The word I have is schoolmistresses.

In another problem I got the equation x+y+5z+w=15 and I got to find out how many solution this equation got. I'm not sure how to do this in itself and especially not with the conditions : x; y; z and w are non-negative integers and we also have that x is positive?
I will help with these two.
The first follows the standard example: consider $MISSISSIPPI$. There are there are eleven letters there: four S's, four I's, and two P's are repeated.
There are $\frac{11!}{(4!)(4!)(2!)}$ ways to rearrange that word.
We divide to eliminate the over counts.

For $x+y+5z+w=15$ the $5$ complicates things.
The z can only four values: $0,1,2,3$

You should know that $x+y+w=15$ has $\binom{15+3-1}{15}$ non-negative integral solutions. So that takes care of $z=0.$

Now for $z=1$, $\binom{10+3-1}{10}$. Why the 10? Because we have already used 1(5)=5 of the ones.

To finish do that for $z=2,3$ and add the four results.

3. ## Re: Combinatorics

Originally Posted by Plato
There is just one thing I dont understand with the last part and that is how you come up with 15+3-1?
Thanks for the help!