Properties of binomial coefficients

The row we work with

The expansion of $(1 + x)^n$ has the form

The numbers ${}^{n}C_{r}$ are the binomial coefficients, and each row of Pascal's triangle lists them. Row $6$ reads $1, 6, 15, 20, 15, 6, 1$. It reads the same from either end.

This lesson turns what you see in a row into five rules. Each one saves a long calculation.

Symmetry

Choosing $r$ things to take is the same as choosing $n - r$ things to leave behind. So

The formula agrees. The fraction $\dfrac{n!}{r!\,(n - r)!}$ does not change when $r$ and $n - r$ swap places.

Use the rule when $r$ is close to $n$. For example, ${}^{12}C_{9} = {}^{12}C_{3} = \dfrac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$, and ${}^{15}C_{13} = {}^{15}C_{2} = 105$.

When two coefficients are equal

Along a row, the coefficients rise to the middle and then fall. So two different places hold the same value only when they are mirror images. That gives a rule.

If ${}^{n}C_{a} = {}^{n}C_{b}$ and $a \ne b$, then $a + b = n$.

Example 1. If ${}^{n}C_{7} = {}^{n}C_{5}$, then $n = 7 + 5 = 12$.

Example 2. If ${}^{20}C_{x} = {}^{20}C_{x + 6}$, the two places are different, so $x + (x + 6) = 20$ and $x = 7$.

Pascal's rule

Add two neighbours in a row and you get the number below them in the next row:

Why it is true. Take $n + 1$ people and choose $r$ of them. Point at one person, Meena. Either Meena is chosen, and $r - 1$ more come from the other $n$ people: ${}^{n}C_{r - 1}$ ways. Or she is not chosen, and all $r$ come from the other $n$: ${}^{n}C_{r}$ ways. The two cases cover everything.

Check. ${}^{9}C_{3} + {}^{9}C_{4} = 84 + 126 = 210 = {}^{10}C_{4}$.

The ratio of neighbours

Divide ${}^{n}C_{r}$ by ${}^{n}C_{r - 1}$. Almost all of the factorials cancel:

So each coefficient is the one before it times $\dfrac{n - r + 1}{r}$. You can build a row without writing a single factorial. For row $8$:

• Start with $1$.
• Times $\tfrac{8}{1}$ gives $8$.
• Times $\tfrac{7}{2}$ gives $28$.
• Times $\tfrac{6}{3}$ gives $56$.
• Times $\tfrac{5}{4}$ gives $70$.

The rest of the row follows by symmetry: $56, 28, 8, 1$.

Move $r$ and watch the two readouts. The ratio of the two coefficients always matches $\dfrac{n - r + 1}{r}$.

The identity for $r \cdot {}^{n}C_{r}$

Why it is true. Choose a team of $r$ people from $n$, and then pick a captain from the team. That is $r \cdot {}^{n}C_{r}$ ways. Count the same thing the other way. Pick the captain first, in $n$ ways, and then choose the other $r - 1$ team members from the remaining $n - 1$ people.

Check. $4 \cdot {}^{9}C_{4} = 4 \times 126 = 504$, and $9 \cdot {}^{8}C_{3} = 9 \times 56 = 504$.

The identity removes a factor of $r$ from a sum. The next lesson uses it.

Three consecutive coefficients

Example. Three consecutive coefficients in $(1 + x)^n$ are in the ratio $1 : 3 : 5$. Find $n$.

Step 1. Name them. Call them ${}^{n}C_{r - 1}$, ${}^{n}C_{r}$ and ${}^{n}C_{r + 1}$.

Step 2. Two ratios. $\dfrac{n - r + 1}{r} = 3$ and $\dfrac{n - r}{r + 1} = \dfrac{5}{3}$.

Step 3. Solve. The first gives $n = 4r - 1$. Put that in the second: $3(3r - 1) = 5(r + 1)$, so $4r = 8$ and $r = 2$. Then $n = 7$.

Check. ${}^{7}C_{1}$, ${}^{7}C_{2}$ and ${}^{7}C_{3}$ are $7$, $21$ and $35$, and $7 : 21 : 35 = 1 : 3 : 5$.

Practise with new numbers

Try it in Python

The program checks symmetry, Pascal's rule, the identity for $r \cdot {}^{n}C_{r}$ and the ratio rule in every row from $1$ to $12$. Then it builds row $8$ with the ratio rule.

Common mistakes

• Matching the wrong place. ${}^{n}C_{r}$ equals ${}^{n}C_{n - r}$, not ${}^{n}C_{n + r}$ and not ${}^{n}C_{r - n}$.
• Forgetting the first case. If ${}^{n}C_{a} = {}^{n}C_{b}$, then $a = b$ or $a + b = n$. Only the second gives new information.
• Adding from two rows. Pascal's rule adds ${}^{n}C_{r - 1}$ and ${}^{n}C_{r}$ from the same row. The answer sits in row $n + 1$.
• Turning the ratio upside down. The next coefficient is the one before times $\dfrac{n - r + 1}{r}$, not $\dfrac{r}{n - r + 1}$.
• Dropping a factor in the identity. On the right side, $n$ multiplies ${}^{n - 1}C_{r - 1}$. The top and the bottom both fall by $1$.

Remember these

• ${}^{n}C_{r} = {}^{n}C_{n - r}$.
• If ${}^{n}C_{a} = {}^{n}C_{b}$ with $a \ne b$, then $a + b = n$.
• ${}^{n}C_{r - 1} + {}^{n}C_{r} = {}^{n + 1}C_{r}$.
• ${}^{n}C_{r} = {}^{n}C_{r - 1} \times \dfrac{n - r + 1}{r}$.
• $r \cdot {}^{n}C_{r} = n \cdot {}^{n - 1}C_{r - 1}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the value of ${}^{10}C_{8}$.

2. If ${}^{n}C_{4} = {}^{n}C_{6}$, find $n$.

3. Use Pascal's rule to find ${}^{7}C_{2} + {}^{7}C_{3}$.

4. In row $8$, ${}^{8}C_{4} = 70$. Use the ratio rule to find ${}^{8}C_{5}$.

5. Show that $6 \cdot {}^{9}C_{6} = 9 \cdot {}^{8}C_{5}$ by finding both values.