Binomial Coefficients

7713 /www/nrich/html/content/id/7713/ 3 0000-00-00T00:00:00 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">This article introduces the Binomial Coefficients. It is best read with paper and pen so that you can have a go at the questions as you read. There are several ways of defining the binomial coefficients, but for this article we will be using the following definition and notation: \[n\choose k\] (pronounced “$n$ choose $k$”) is the number of distinct subsets of size $k$ of a set of size $n$. More informally, it's the number of different ways you can choose $k$ things from a collection of $n$ of them (hence $n$ choose $k$). So: <ul> <li> ${3 \choose 1} = 3$ because you can either pick the first thing, or the second thing, or the third thing. </li> <li> Likewise, ${3 \choose 2} = 3$ because you can either take the first two, the first and the third, or the last two. Notice that the order you choose the things in is ignored. </li> </ul> Can you work out the following binomial coefficents? <ol style="list-style-type: decimal"> <li>$4 \choose 1$</li> <li>$4 \choose 2$</li> <li>$5 \choose 2$</li> <li>$6 \choose 2$</li> </ol> You'll need to come up with a systematic method of making sure you've found all the ways of choosing. Does your method suggest to you any way of calculating the result directly? Now go back to the definition in terms of choosing things, and see if you can work out why the following are true for all $n\ge 1$: \[{n \choose 0} = {n \choose n} = 1 \] \[{n \choose 1} = {n \choose n-1} = n \] This starts to suggest a pattern, in fact: \[{n \choose k} = {n \choose n-k}\] Can you justify this pattern from the definition? Now let's write out the binomial coefficients in a grid: <table> <tbody> <tr> <th> </th> <th>0</th> <th>1</th> <th>2</th> <th>3</th> <th>4</th> </tr> <tr> <th>0</th> <td>1</td> <td> </td> <td> </td> <td> </td> <td> </td> </tr> <tr> <th>1</th> <td>1</td> <td>1</td> <td> </td> <td> </td> <td> </td> </tr> <tr> <th>2</th> <td>1</td> <td>2</td> <td>1</td> <td> </td> <td> </td> </tr> <tr> <th>3</th> <td>1</td> <td>3</td> <td>3</td> <td>1</td> <td> </td> </tr> <tr> <th>4</th> <td>1</td> <td>4</td> <td>6</td> <td>4</td> <td>1</td> </tr> </tbody> </table> Do these <a href="/5347">look familiar</a>? They should do. This suggests the following rule: \[{n \choose k} = {n-1 \choose k} + {n-1 \choose k-1}\] Thinking back to your systematic method, can you explain this relation in terms of choosing things? We can use this relation to calculate binomial coefficients, but it's not very efficient. You'll find yourself having to calculate more and more binomial coefficients the larger your $n$ and $k$. So let's try to find another formula, by thinking about the process of choosing: <ol style="list-style-type: decimal"> <li> Suppose you want to choose five things from a collection of twelve. How many choices do you have for the first thing? How many choices for the second? How many choices for the third, fourth, and fifth? How many choices does this give altogether? </li> <li> The problem with the above method is that it counts 1,2,3,4,5 and 2,1,3,4,5 separately even though they are really the same choice. How many times do you count each choice? How can you eliminate them from your count? </li> </ol> Once you've come up with a formula, can you use it to justify the earlier results algebraically? <h3 id="the-connection-with-the-binomial">The connection with the binomial</h3> I mentioned these were called binomial coefficients at the beginning of the article, but I haven't mentioned the binomial formula since. Well, the binomial formula is this: \[(a+b)^n = \sum_{k=0}^n {n \choose k} a^k b^{n-k}\] So they are literally just the coefficients of $a^k$ in the expansion of the $n^\textrm{th}$ power. Thinking about $(a+b)^n$ as $(a+b)(a+b)(a+b)\dots(a+b)$ and expanding in the usual way, can you see how the choosing-from-sets definition is connected to this idea? <h3 id="extensions">Extensions</h3> These formulae all have justifications both in terms of choosing things and in terms of algebra. See if you can find one or the other (or both!): \[\sum_{k=0}^{n-r} {n-k \choose r} = {n+1 \choose r+1}\] \[\sum_{k=0}^n {n \choose k} = 2^n\] \[\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}\]</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> When trying to work out the number of ways of choosing things, you'll first need to come up with a good way of keeping track of what choices you've found already. You could write T for each thing you took and L for each thing you left, and represent choosing 2 from 5 as TTLLL, TLTLL, TLLTL, ... You could decide you were choosing from the first $n$ natural numbers, and represent choosing 2 from 5 as {1,2}, {1,3}, {1,4}, ... You should decide which representation feels most natural to you. Which is the most useful may depend on the problem you're trying to solve at the time, so try both. To be sure you have all the choices, look for patterns. Ask yourself, do I have all the choices where I take the first thing? What about all the choices where I don't take the first thing? When you're trying to prove that ${n \choose k} = {n \choose n-k}$, consider why ${3 \choose 1}$ turned out to be the same as ${3 \choose 2}$. What would be a fast way to choose 99 elements from 100? For the addition rule, I think the TLTLL representation is more useful than the {1,3} representation. How many choices begin with T? How many begin with L? For the first extension, one approach is to just apply the addition rule repeatedly. If you had a choices interpretation of the addition rule, think about what that means here. This time the {1,3} representation is probably more useful. For the second extension, consider what the left-hand side means in terms of choices, or try a proof by induction, or try using the binomial formula. For the third extension, we're trying to write a choice from $2n$ things in terms of choices from $n$ things. How could you choose from $2n$ things by making choices from sets of $n$ things? Alternatively, try using the binomial formula: see also <a href="/350">Binomial</a>. These extensions are meant to be very hard, so don't lose sleep over them. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <ul> <li>${4 \choose 1} = 4$</li> <li>${4 \choose 2} = 6$</li> <li>${5 \choose 2} = 10$</li> <li>${6 \choose 2} = 15$</li> </ul> \[{n \choose 0} = {n \choose n} = 1 \] is true because there's only one way of choosing no things, or all the things. \[{n \choose 1} = {n \choose n-1} = n \] is true because either we're including a single element, or precluding a single element, and there are $n$ such elements to include or preclude. \[{n \choose k} = {n \choose n-k}\] is true because choosing $k$ elements to take from $n$ is equivalent to choosing $n-k$ elements to leave behind. \[{n \choose k} = {n-1 \choose k} + {n-1 \choose k-1}\] is true because we can either leave the first element and choose $k$ from the remaining $n-1$, or we can take the first element, in which case we still need $k-1$ more from the remaining $n-1$. Deriving the formula: <ol style="list-style-type: decimal"> <li> There are $n \times (n-1) \times \dots \times (n-k+1)$ choices altogether. </li> <li> Each choice is counted $k!$ times because there are $k!$ orderings of $k$ elements ($k$ choices for the first, $k-1$ choices for the second...) and each ordering is counted exactly once. We compensate by dividing by $k!$. </li> </ol> So we get \[\frac{\prod_{r=n-k+1}^n r}{k!}\] which can be more simply written as \[\frac{n!}{k!(n-k)!}\] Then the symmetry between $k$ and $n-k$ follows since $n-(n-k)=k$, the results with $k$ being 0 and 1 and hence $n$ and $n-1$ follow by substitution, and the addition formula follows like so: \[\begin{align*} {n-1 \choose k-1} + {n-1 \choose k} &= \frac{(n-1)!}{(k-1)!(n-k)!} + \frac{(n-1)!}{k!(n-k-1)!} \\ &= \frac{(n-1)!}{(k-1)!(n-k)(n-k-1)!} + \frac{(n-1)!}{k(k-1)!(n-k-1)!} \\ &= \frac{(n-1)!}{(k-1)!(n-k-1)!}\left(\frac{1}{n-k} + \frac{1}{k}\right) \\ &= \frac{(n-1)!}{(k-1)!(n-k-1)!}\left( \frac{k + (n-k)}{k(n-k)}\right) \\ &= \frac{(n-1)!n}{(k-1)!(n-k-1)!k(n-k)} \\ &= \frac{n!}{k!(n-k)!} \\ &= {n \choose k} \end{align*}\] The binomial coefficients arise because in the expansion of $(a+b)(a+b)\dots(a+b)$ we get one $a^k b^{n-k}$ for each set of $k$ $a$'s we can pick out from the $n$ brackets. <h3 id="extensions">Extensions</h3> The first sum can be algebraically shown by repeated application of the addition rule above: \[\begin{align*} {n \choose r} &= {n-1 \choose r-1} + {n-1 \choose r} \\ &= {n-1 \choose r-1} + {n-2 \choose r-1} + {n-2 \choose r} \\ &= \dots \end{align*}\] continuing until we get to ${r-1 \choose r} = 0$. Then we make the substition $n' = n-1$ and $r' = r-1$ to make it a bit neater. Alternatively, we're picking $r+1$ of the numbers $\{1\dots n+1\}$. Let's pick our smallest number first to be $k+1$, and then pick the remaining $r$ numbers from the $n-k$ larger numbers. Once we've done this for every possible $k$ we've made all the possible choices. The second sum adds up the number of ways of choosing any number of elements from $n$ elements. To work out how many this is, just go through each element at a time and decide to take it or leave it: two choices for each element is $2^n$ choices. Alternatively, go back to Pascal's Triangle: apply the addition rule to every term of the sum on the LHS (just define the awkward ones to be zero). You should find that you get every term on the previous row exactly twice each, so the row sums double every time, so you get $2^n$. Alternatively, expand $(a+b)^n$ with the binomial formula and then set $a=b=1$. For the third sum, have a look at <a href="/350/solution">Binomial</a>. Alternatively, take $2n$ things and cut them into two groups of size $n$. Then you choose 0 from one and $n$ from the other, or 1 from one and $n-1$ from the other, etc. to get: \[{2n \choose n} = {n \choose 0}{n \choose n} + {n \choose 1}{n \choose n-1} + \dots\] Then apply the symmetry formula to see that ${n \choose k}{n \choose n-k} = {n \choose k}^2$. </mdoxml> 2 4 0 0 0 1 1 Binomial Coefficients An introduction to the binomial coefficient, and exploration of some of the formulae it satisfies. 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