Choosing a committee of 3 from 10 people is the same as choosing the 7 people who are left out, so $^{10}C_3 = \,{}^{10}C_7$. This symmetry is one of the most powerful shortcuts in combinatorics. In this lesson you'll lock in three key properties of $^nC_r$ and discover Pascal's identity, the rule that builds the entire triangle from just $1$s.
You need to choose 2 people from a group of 5 to be on a committee. You also need to choose 3 people to be left off. Without calculating do you think $^5C_2$ equals $^5C_3$? Why or why not? Write your gut answer before reading on.
Every combinations property question reduces to two operations: use the symmetry rule to rewrite $^nC_r = \,{}^nC_{n-r}$, or apply Pascal's identity to combine adjacent entries into the row below.
Choosing $r$ objects from $n$ is identical to choosing the $n-r$ objects to leave out. Both choices define the same subset. Pascal's identity then says: any entry in the triangle equals the sum of the two entries directly above it.
Every time you choose $r$ objects from $n$, you simultaneously determine the $n - r$ objects that are left out. These two choices are in perfect one-to-one correspondence, so the number of ways to do each is equal:
Algebraic verification:
$^nC_{n-r} = \dfrac{n!}{(n-r)!\,(n-(n-r))!} = \dfrac{n!}{(n-r)!\,r!} = \,{}^nC_r$ ✓
Examples:
Symmetry: ^nC_r = \,^nC_{n-r}, choosing r items and choosing the n-r items to leave out are the same count; Use symmetry to simplify: always compute ^nC_k using the smaller of k and n-k
Pause, copy the symmetry identity into your book: $^nC_r = ^nC_{n-r}$, choosing $r$ from $n$ equals choosing $n-r$ from $n$; use it to reduce computation by always taking the smaller of $r$ and $n-r$.
Quick check: Which expression equals $^{15}C_{12}$?
We just saw that $^nC_r = ^nC_{n-r}$, choosing $r$ to include is the same count as choosing $n-r$ to exclude. That raises a question: the symmetry identity connects adjacent rows, but is there a rule that combines two adjacent entries in the same row to give an entry in the next row? This card answers it → Pascal's identity: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$; this is why each entry in Pascal's triangle equals the sum of the two above it.
Pascal's identity is the rule that generates each row of Pascal's triangle from the one above it:
Combinatorial explanation: Suppose we want to choose $r$ people from a group of $n+1$, including a particular person called Alex. Either Alex is chosen (then we need $r-1$ more from the remaining $n$: giving $^nC_{r-1}$ ways) or Alex is not chosen (then we choose all $r$ from the remaining $n$: giving $^nC_r$ ways). Adding both cases:
This is exactly Pascal's identity. The identity is not just a formula, it is a counting argument.
Pascal's identity: $^nC_{r-1} + \,{}^nC_r = \,{}^{n+1}C_r$. The entry in row $n+1$ equals the sum of its two parents in row $n$.
Pascal's identity: ^nC_r + \,^nC_{r-1} = \,^{n+1}C_r; Reading it: add any two adjacent entries in row n to get the entry between them in row n+1
Pause, copy Pascal's identity into your book: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$; reading it, adding any two adjacent entries in row $n$ gives the entry between them in row $n+1$.
Did you get this? True or false: $^7C_3 + \,{}^7C_4 = \,{}^8C_4$.
Worked examples · 3 in a row, reveal as you go
Verify Pascal's identity for $n = 5$, $r = 2$. That is, confirm that $^5C_2 + \,{}^5C_1 = \,{}^6C_2$.
Simplify $^{n+1}C_3 - \,{}^nC_3$.
Without a calculator: (a) Evaluate $^{12}C_{10}$. (b) Find the value of $n$ if $^nC_0 + \,{}^nC_1 + \cdots + \,{}^nC_n = 512$.
Fill the gap: By symmetry, $^{10}C_8 = \,{}^{10}C_{\,}$ $= 45$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $^nC_r + \,{}^nC_{r+1} = \,{}^{n+1}C_r$ is a correct statement of Pascal's identity.
Activities · practice with the ideas
Without a calculator, evaluate $^{13}C_{11}$ using the symmetry property. Show your working.
Verify Pascal's identity for $n = 4$, $r = 2$. Calculate each side independently.
Simplify $^nC_2 + \,{}^nC_1$ using Pascal's identity. State your final answer as a single combination.
A set has $n$ elements and $64$ subsets in total. Find $n$. (Hint: use the row sum formula.)
Explain in your own words why $^nC_r = \,{}^nC_{n-r}$ using a real-world example involving choosing a committee.
Earlier you were asked: do you think $^5C_2$ equals $^5C_3$?
Yes: $^5C_2 = \dfrac{5 \times 4}{2} = 10$ and $^5C_3 = \dfrac{5 \times 4 \times 3}{6} = 10$. They are equal. Choosing 2 people from 5 is the same as choosing the 3 who are left out, they determine each other uniquely. This symmetry, $^nC_r = \,{}^nC_{n-r}$, means the rows of Pascal's triangle read identically from left to right and from right to left.
Pascal's identity $^nC_r + \,{}^nC_{r-1} = \,{}^{n+1}C_r$ is even more powerful: it shows you can build the entire triangle from just the two boundary $1$s, adding neighbours to produce the next row forever. The key insight is that these are not arbitrary arithmetic facts, they reflect deep counting logic.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Evaluate $^{12}C_5$ without a calculator. Show all working. (2 marks)
Q2. Verify Pascal's identity for $n = 6$, $r = 3$. That is, show that $^6C_3 + \,{}^6C_2 = \,{}^7C_3$. (2 marks)
Q3. (a) Simplify $^nC_2 + \,{}^nC_1$ using Pascal's identity. (b) Hence find the value of $^{50}C_2 + \,{}^{50}C_1$ without expanding. (c) Explain why $^nC_r + \,{}^nC_{r-1}$ always produces a value that is larger than either term alone (for $1 \leq r \leq n-1$). (3 marks)
Activity answers: 1. $^{13}C_{11} = \,{}^{13}C_2 = \frac{13 \times 12}{2} = 78$ · 2. $^4C_2 + \,{}^4C_1 = 6 + 4 = 10 = \,{}^5C_2$ ✓ · 3. $^nC_2 + \,{}^nC_1 = \,{}^{n+1}C_2$ · 4. $2^n = 64 = 2^6$, so $n = 6$ · 5. Choosing 2 people for committee is same as choosing 3 to leave out, both select the same partition of the group.
Q1 (2 marks): $^{12}C_5 = \dfrac{12 \times 11 \times 10 \times 9 \times 8}{5!} = \dfrac{95\,040}{120} = 792$ [2 marks: 1 for correct numerator product, 1 for final answer].
Q2 (2 marks): $^6C_3 = 20$, $^6C_2 = 15$; LHS $= 20 + 15 = 35$. $^7C_3 = \frac{7 \times 6 \times 5}{6} = 35$ ✓ [1 mark each side].
Q3 (3 marks): (a) $^nC_2 + \,{}^nC_1 = \,{}^{n+1}C_2$ by Pascal's identity with $r = 2$ [1]. (b) $^{50}C_2 + \,{}^{50}C_1 = \,{}^{51}C_2 = \frac{51 \times 50}{2} = 1275$ [1]. (c) Both terms are positive for $1 \leq r \leq n-1$, so their sum exceeds either term alone; alternatively, $^{n+1}C_r > \,{}^nC_r$ and $^{n+1}C_r > \,{}^nC_{r-1}$ since we are choosing from a larger set [1].
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