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Section 4.4 Binomial Identities

Each of the following problems provides a specific problem modeling an identity involving binomial coefficients.

Problem 4.4.1.

Acme corporation is selecting members for a user feedback panel. The panel will consist of \(k\) characters from a pool of \(n\) characters.

  1. How many ways can \(k\) be selected from \(n\text{?}\)

  2. How many ways can \(k\) be selected if Wile E. Coyote is one of them?

  3. How many ways can \(k\) be selected if Wile E. Coyote is not one of them?

  4. State and explain an identity relating part 1 to parts 2 and 3. Note this is one proof of the identity \({n \choose k}={n-1 \choose k}+{n-1 \choose k-1}\text{.}\)

  5. Prove this identity using algebra.

Solution.

\(\binom{n}{k} = \frac{n!}{k!(n-k)!}\text{.}\) If Wile E. Coyote is already in the set, then (n-1) choose (k-1). We are not choosing him because he has already been selected.

\begin{equation*} {n-1 \choose k-1} = \frac{(n-1)!}{(k-1)!((n-1)-(k-1))!} = \frac{(n-1)!}{(k-1)!(n-k)!} \end{equation*}

If Wile E. Coyote is not in the set, then (n-1) choose k.

\begin{equation*} {n-1 \choose k} = \frac{(n-1)!}{k!((n-1)-k)!} = \frac{(n-1)!}{k!(n-1-k)!} \end{equation*}

For all possible choices, add both together, to get the universal set.

\begin{align*} \frac{(n-1)!}{(k-1)!(n-k)!} + \frac{(n-1)!}{(k!(n-1-k)!)}\\ & = (n-1)!\left(\frac{1}{(k-1)!(n-k)!} + \frac{1}{k!(n-1-k)!}\right)\\ & = (n-1)!\left(\frac{k}{k(k-1)!(n-k)!} + \frac{(n-k)}{(k!(n-k)(n-k-1)!)}\right)\\ & = (n-1)!\left(\frac{k}{k!(n-k)!} + \frac{(n-k)}{k!(n-k)!}\right)\\ & = (n-1)!\left(\frac{n}{k!(n-k)!}\right)\\ & = \frac{n!}{k!(n-k)!}\\ & = {n \choose k} \end{align*}

Problem 4.4.2.

Allen and Betty have different management styles. Allen selects \(k\) employees out of \(n\) and then selects \(m\) leaders from this group of \(k\text{.}\) Betty selects \(m\) leaders out of the \(n\) and allows them to select the rest of the \(k\) employees from those remaining.

  1. How many ways can Allen select his way?

  2. How many ways can Betty select her way?

  3. State and explain a relation between the previous. Note this is one proof of the identity \({n \choose k} {k \choose m} = {n \choose m} {n-m \choose k-m}\text{.}\)

  4. Prove the relation using algebra.

Solution.

1) n choose k times k choose m, or \(\frac{n!}{k! (n - k)!)} \cdot \frac{k!}{m! (k - m)!)}\)

From n people total, choose k employees. Then for each way this can be achieved, choose m leaders out of k employees.

2) n choose m times \(n - m\) choose \(k - m\text{,}\) or \(\frac{n!}{m! (n - m)!)} \cdot \frac{(n - m)!}{(k - m)! (n - m - (k - m))!)}\)

This can be rewritten: \(\frac{n!}{m! (n - m)!)} \cdot \frac{(n - m)!}{(k - m)! (n - k))!)}\)

From n people total, choose m leaders. Then for each way this can be achieved, select from the remaining \(n - m\) people enough employees to total k employees including the m leaders we have. This is \(k - m\) employees to select from \(n - m\)

3) The ways Allen and Betty select employees have the same result. The only difference is the order — Allen first chooses k employees, then chooses m leaders from the k, and Betty first chooses m leaders, then adds more employees to make k total. In both cases, n is the total number of people, there are m leaders and k employees, and we did not adjust any of these numbers for either method.

We will prove that the two methods are equal by beginning with the second method and showing that it is equal to the first method using algebra.

Problem 4.4.3.

“Block Walking” is a model for binomial coefficients. Use the provided block walking diagram for the following problems. Coordinates below are \((n,k)\) where \(n\) is the position vertically (top to bottom) and \(k\) is the position (block) along the right diagonal.

  1. Write down all the ways to get from \((0,0)\) to \((4,2)\) by recording whether one walks left or right from each intersection.

  2. Explain how the number of ways to get from \((0,0)\) to \((4,2)\) could be counted without listing all of the routes.

Problem 4.4.4.

Use the block walking model for the following problems.

  1. Count the number of right moves necessary to get from \((0,0)\) to \((4,2)\) that start by going left first.

  2. Count the number of additional right moves necessary to get from \((0,0)\) to \((4,2)\) that start by going right first.

  3. Explain how these two counts relate to the total number of ways to get from \((0,0)\) to \((4,2)\text{.}\)

  4. Use this to prove the identity \({ n \choose k } = { n-1 \choose k} + { n-1 \choose k-1}\text{.}\)

Problem 4.4.5.

Prove the following identities using the block walking method.

  1. \({n \choose 0}+{n \choose 1}+{n \choose 2}+\ldots+{n \choose n} = 2^n\text{.}\)

  2. \({n \choose 0}+{n+1 \choose 1}+{n+2 \choose 2}+\ldots+{n+r \choose r} = {n+r+1 \choose r}\text{.}\)

  3. \({r \choose r}+{r+1 \choose r}+{r+2 \choose r}+\ldots+{n \choose r} = {n+1 \choose r+1}\text{.}\)

  4. \({n \choose 0}^2+{n \choose 1}^2+{n \choose 2}^2+\ldots+{n \choose n}^2 = {2n \choose n}\text{.}\)

  5. \(\displaystyle \displaystyle \sum_{k=0}^r {m \choose k}{n \choose r-k} = {m+n \choose r}\)

  6. \(\displaystyle \displaystyle \sum_{k=0}^m {m \choose k}{n \choose r+k} = {m+n \choose m+r}\)

  7. \(\displaystyle \displaystyle \sum_{k=n-s}^{m-r} {m-k \choose r}{n+k \choose s} = {m+n+1 \choose r+s+1}\)