# Combinatorics Count

Can someone help me explain a problem where suppose you have n antenna of which m are defective and n-m are functional and the functional antenna would be between two defective ones then the no of non negative integer would be n-m+2. So the no of possible ways that can be arranged would (n-m+1 C m) Similarly if there is two functional antenna between two defective ones then the no of non negative integer would be n-2m+3. This part is from Sheldon Ross book chapter 1. Please help me in explaining this?

Manish Agrawal

-
I understood that thing but now I am unable to understand why there is n-m+2 and n-2m+3 in the result. Book says that assume y1 = x+1 and yi = xi where i = 2,3,4,5....n and ym+1 = xm+1 then the non negative integer would be n-m+2..how this is happening – user86354 Jul 18 '13 at 12:31
I'm finding it hard to understand this question (e.g. what does "no of non negative integer would be n-m+2" mean?). If this is asking about Example 4c from the book, then there is a worked solution in the book; I'm unlikely to be able to provide a substantially better explanation. – Douglas S. Stones Jul 18 '13 at 12:38
This is from the last page of the chapter..the last example... – user86354 Jul 18 '13 at 12:52

We solve the second problem, where there are always at least two goods between any two bads. A simpler version of the same idea takes care of the first problem. (The first problem, under various wordings, has appeared often on MSE.)

We try to give a visual argument. Temporarily, we change the notation. Let $b$ be the number of bad antennas, and let $g$ be the number of good antennas. Note that $b$ is called $m$ in the problem, and $g$ is called $n-m$.

Because we will be drawing a diagram, it is useful to use concrete values of $b$ and $g$. So suppose that $b=6$ and $g=14$.

Make $5$ bundles by putting a good antenna immediately to the right of all but one of the bad antennas. We leave that one untouched. So now we have the following set of objects in the truck: $$\{BG, BG, BG, BG, BG,G\}.$$ In symbols, we have used $b-1$ goods to make the bundles. This leaves us $g-(b-1)=g-b+1$ good antennas, in our case $9$. Line up these $9$ good antennas like this: $$G \qquad G \qquad G \qquad G \qquad G \qquad G \qquad G \qquad G\qquad G$$

These $9$ $G$'s determine $8$ gaps. But there are also the two ends, which we will call endgaps. So we have $10$ "gaps" to put our objects into.

We need to choose $6$ of these "gaps." The rule will be that the lone $B$ will go into the rightmost chosen gap. The number of ways to choose $6$ gaps from $10$ is $\binom{10}{6}$.

In general, there will be $g-b+2$ "gaps" and we must choose $b$ of them. So the number of ways to do the job is $$\binom{g-b+2}{b}.\tag{1}$$

Now recall that $b=m$ and $g=n-m$. Substituting in (1) we find that the number of arrangements is $$\binom{n-2m+2}{m}.$$

Remark: The argument can be rephrased in terms of expressing an integer as a sum of positive integers. An alternative argument uses the bundling $\{BGG,BGG,BGG,BGG,BGG,B\}$.

-
Awesome explanation!!! Thank you very much. – user86354 Jul 18 '13 at 18:07