Course 4.3Lesson 2 of 3

Properties of nCr

Choosing $r$ to take is choosing $n - r$ to leave, so $^{n}C_{r} = {}^{n}C_{n - r}$

about 25 min1 simulation1 practice question1 Python program5 questions to check yourself

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Taking or leaving

Rahim packs $3$ of his $5$ shirts for a trip. Each choice of $3$ to take is also a choice of $2$ to leave behind. So the two counts are equal:

\displaystyle ^{5}C_{3} = {}^{5}C_{2} = 10.

In general, choosing $r$ to take is choosing $n - r$ to leave:

\displaystyle ^{n}C_{r} = {}^{n}C_{n - r}.

The formula shows it too. Swap $r$ and $n - r$ in $\dfrac{n!}{r!\,(n - r)!}$ and nothing changes. This saves work: $^{20}C_{18} = {}^{20}C_{2} = \dfrac{20 \times 19}{2} = 190$ .

Two equal values

If $^{n}C_{a} = {}^{n}C_{b}$ , then either $a = b$ or $a + b = n$ . So when $a \ne b$ , you can read off $n$ .

For example, if $^{n}C_{9} = {}^{n}C_{7}$ , then $n = 9 + 7 = 16$ . And then $^{n}C_{2} = {}^{16}C_{2} = 120$ .

Pascal's rule

\displaystyle ^{n}C_{r} + {}^{n}C_{r - 1} = {}^{n + 1}C_{r}, \qquad 1 \le r \le n.

A counting proof. A class of $n + 1$ students includes Ishaan. Count the groups of $r$ from the class in two ways.

Groups without Ishaan: choose all $r$ from the other $n$ , in $^{n}C_{r}$ ways.
Groups with Ishaan: choose his $r - 1$ partners from the other $n$ , in $^{n}C_{r - 1}$ ways.

Every group is in exactly one of the two kinds. So the total, $^{n + 1}C_{r}$ , is the sum.

An algebra proof is in the problems. It puts both fractions over the common bottom $r!\,(n - r + 1)!$ .

Pascal's triangle

Write the values of $^{n}C_{r}$ in rows, one row for each $n$ . Each row starts and ends with $1$ . By Pascal's rule, every other number is the sum of the two just above it.

$n$	the row
$4$	$1 \quad 4 \quad 6 \quad 4 \quad 1$
$5$	$1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1$
$6$	$1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1$

Each row reads the same from both ends. That is the symmetry $^{n}C_{r} = {}^{n}C_{n - r}$ .

Move $r$ across the row. The values rise to the middle and then fall. The two dots always match.

Counting in geometry

Lines. Take $n$ points, no three on a straight line. Any two of them fix a line. So there are $^{n}C_{2}$ lines.

Triangles. Any three of the points make a triangle. So there are $^{n}C_{3}$ triangles.

Diagonals. A polygon with $n$ corners has $^{n}C_{2}$ segments joining two corners. Of these, $n$ are sides. So it has $^{n}C_{2} - n = \dfrac{n(n - 3)}{2}$ diagonals. A hexagon has $15 - 6 = 9$ .

Points on a line. Now let $4$ of $7$ points lie on one line, with no other three in a line. Three points on that line make no triangle. So the triangles number $^{7}C_{3} - {}^{4}C_{3} = 35 - 4 = 31$ .

Two pictures. Left: six points at the corners of a regular hexagon, joined in every way: 6 blue sides and 9 dashed orange diagonals, 15 segments in all, which is 6C2. Right: seven points, four red ones on a straight line and three green ones above it. One triangle is shaded, with two corners on the line and one above it. The title says the number of triangles is 7C3 minus 4C3, which is 35 minus 4, or 31.

A worked example

If $^{n}C_{r - 1} = 36$ , $^{n}C_{r} = 84$ and $^{n}C_{r + 1} = 126$ , find $n$ and $r$ .

Step 1. A handy ratio. Divide the two formulas: $\dfrac{^{n}C_{r}}{^{n}C_{r - 1}} = \dfrac{n - r + 1}{r}$ .

Step 2. First ratio. $\dfrac{84}{36} = \dfrac{7}{3}$ , so $3(n - r + 1) = 7r$ , or $3n + 3 = 10r$ .

Step 3. Second ratio. $\dfrac{126}{84} = \dfrac{3}{2} = \dfrac{n - r}{r + 1}$ , so $2n = 5r + 3$ .

Step 4. Solve. From Step 3, $n = \dfrac{5r + 3}{2}$ . Then $\dfrac{3(5r + 3)}{2} + 3 = 10r$ , so $15r + 15 = 20r$ and $r = 3$ , $n = 9$ .

Check: $^{9}C_{2} = 36$ , $^{9}C_{3} = 84$ and $^{9}C_{4} = 126$ .

Practise with new numbers

Try it in Python

The program prints Pascal's triangle with comb, then checks the symmetry and Pascal's rule for every row up to $12$ .

Common mistakes

Missing the second answer. $^{n}C_{a} = {}^{n}C_{b}$ allows $a + b = n$ , not only $a = b$ .
Adding the wrong pair. Pascal's rule adds two neighbours in the same row: $^{n}C_{r - 1}$ and $^{n}C_{r}$ .
Counting sides as diagonals. Take the $n$ sides away from $^{n}C_{2}$ .
Forgetting points on a line. Points on one line give one line, not many, and no triangles among themselves.

Remember these

$^{n}C_{r} = {}^{n}C_{n - r}$ .
$^{n}C_{a} = {}^{n}C_{b}$ means $a = b$ or $a + b = n$ .
$^{n}C_{r} + {}^{n}C_{r - 1} = {}^{n + 1}C_{r}$ , Pascal's rule.
$\dfrac{^{n}C_{r}}{^{n}C_{r - 1}} = \dfrac{n - r + 1}{r}$ .
$n$ points, no three in a line: $^{n}C_{2}$ lines and $^{n}C_{3}$ triangles. A polygon of $n$ sides has $\dfrac{n(n - 3)}{2}$ diagonals.