Pascal's Triangle & Its Properties
One triangle. Countless patterns. Every row sums to a power of 2. Every entry can be found with a single combination. Diagonal "hockey-sticks" sum to entries at their tip. In this lesson you'll master the structure of Pascal's triangle and use its properties to answer questions that would otherwise require long calculations.
Row 0 of Pascal's triangle sums to 1. Row 1 sums to 2. Row 2 sums to 4. Row 3 sums to 8. Without calculating — what does row 6 sum to? Is there a pattern, and if so, what is it? Write your gut answer before reading on.
Every Pascal's triangle question reduces to two operations: identify the row and position (numbering both from zero), then read off the entry as $^nC_k$ or apply a property (row sum, hockey-stick, symmetry) to answer without building the full triangle.
Row $n$ of Pascal's triangle contains the values $^nC_0, \,^nC_1, \ldots, \,^nC_n$. The top row is row 0, the leftmost entry of each row is position 0. Both start at zero — a perpetual exam trap for students who number from 1.
Key facts
- Row $n$ contains $^nC_0, \,^nC_1, \ldots, \,^nC_n$; both row and position number from 0
- Row $n$ sum $= 2^n$
- Hockey-stick identity: $\sum_{k=r}^{n}\,^kC_r = \,^{n+1}C_{r+1}$
Concepts
- Why each row is a palindrome (symmetry of combinations)
- Why the row sum equals $2^n$ (connection to subsets)
- Why the hockey-stick sums a diagonal into a single entry
Skills
- Write out any row of Pascal's triangle using $^nC_k$
- Find any entry by row and position, or use properties to avoid building the triangle
- Apply the hockey-stick identity to evaluate diagonal sums
Pascal's triangle is an infinite triangular array where row $n$ contains the values $^nC_0, \,^nC_1, \ldots, \,^nC_n$. The top entry (row 0) is $^0C_0 = 1$. Each subsequent row is built by placing $1$s at both ends and using Pascal's identity for every interior entry.
Rows 0–5 of Pascal's triangle. Each row $n$ sums to $2^n$ (shown right). Both rows and positions count from zero.
The entry at row $n$, position $k$ (both starting from 0) is always $^nC_k$. To find an entry, you need only two pieces of information: the row number and the position within that row.
Row n contains ^nC_0, \,^nC_1, , \,^nC_n — that is n+1 entries in total; Numbering from zero: top row is row 0, leftmost entry is position 0
Pause — copy Pascal's triangle structure into your book: row $n$ has $n+1$ entries — $^nC_0, ^nC_1, \ldots, ^nC_n$; rows numbered from zero; each entry = sum of two above (Pascal's identity); leftmost and rightmost entries always equal 1.
Quick check: In Pascal's triangle, the top row (containing just "1") is called row number:
We just saw that Pascal's triangle row $n$ contains $^nC_0, ^nC_1, \ldots, ^nC_n$ with each entry the sum of the two above it (Pascal's identity). That raises a question: beyond its construction rule, Pascal's triangle has properties about row sums and symmetry — what are these, and how do you prove them? This card answers it → row sum $= 2^n$ (set $a=b=1$ in Binomial Theorem); symmetry $^nC_k = ^nC_{n-k}$ (rows are palindromes).
Beyond its construction rule (each entry = sum of two above), Pascal's triangle has three major properties you must know for the HSC.
Property 1 — Row sum:
Proof: set $a = b = 1$ in the Binomial Theorem $(a+b)^n = \sum_{k=0}^{n}\,^nC_k a^{n-k}b^k$. The left side becomes $2^n$; the right side is $\sum\,^nC_k$.
Property 2 — Symmetry: Each row reads the same from left to right and right to left, because $^nC_k = \,^nC_{n-k}$. Row 4: $1, 4, 6, 4, 1$ — a palindrome.
Property 3 — Hockey-stick identity:
Start at $^rC_r = 1$ (the boundary entry), then sum down the diagonal — the total equals the entry one step to the right in the next row down. The name "hockey stick" comes from the shape traced in the triangle: a straight diagonal handle and a curved tip.
Row sum: _{k=0}^n \,^nC_k = 2^n — set a = b = 1 in Binomial Theorem; Symmetry: ^nC_k = \,^nC_{n-k} — rows are palindromes
Pause — copy the three Pascal's triangle properties into your book: (1) row sum $\sum_{k=0}^n ^nC_k = 2^n$ (set $a=b=1$ in Binomial Theorem); (2) symmetry $^nC_k = ^nC_{n-k}$; (3) the leading diagonal contains the triangular numbers.
Did you get this? True or false: the sum of all entries in row 8 of Pascal's triangle is 256.
Worked examples · 3 in a row, reveal as you go
Find the sum of all entries in row 6 of Pascal's triangle.
Find the 4th entry in row 7 of Pascal's triangle (where positions are numbered starting from 0).
Use the hockey-stick identity to evaluate $^2C_2 + \,^3C_2 + \,^4C_2 + \,^5C_2 + \,^6C_2$.
Fill the gap: The middle entry of row 8 (position 4 of row 8) equals $^8C_4 = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the sum of row 10 of Pascal's triangle is 1024.
Odd one out: Which of these is NOT a true property of Pascal's triangle?
Activities · practice with the ideas
Write out row 5 of Pascal's triangle. State each entry as both a number and a combination $^nC_k$.
Find the sum of all entries in row 8. Use the formula — do not add them individually.
Find the middle entry of row 8. (Recall: row 8 has 9 entries, so the middle is position 4.)
Use the hockey-stick identity to evaluate $^3C_3 + \,^4C_3 + \,^5C_3 + \,^6C_3$. Identify $r$ and $n$, state the identity, then compute the answer.
Explain in your own words why the row sum formula $\sum_{k=0}^n \,^nC_k = 2^n$ makes sense in terms of subsets. How many subsets does a set of $n$ elements have?
Earlier you were asked: what does row 6 sum to?
The answer is $2^6 = 64$. The pattern you may have noticed — each row sum doubles the previous one — is exactly right. Row 0: $1 = 2^0$. Row 1: $2 = 2^1$. Row 2: $4 = 2^2$. And so on. This is not a coincidence: it reflects the fact that a set of $n$ elements has exactly $2^n$ subsets, and each subset is counted exactly once somewhere in row $n$.
The triangle is also symmetric (palindrome rows) and obeys the hockey-stick identity. Together, these three properties — row sum, symmetry, and hockey-stick — let you answer most Pascal's triangle questions in seconds, without building the entire triangle from scratch.
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. Write out row 5 of Pascal's triangle, listing all six entries as numbers. (1 mark)
Q2. (a) State the sum of all entries in row 9. (b) Find the entry at row 9, position 2. (2 marks)
Q3. (a) Use the hockey-stick identity to evaluate $^1C_1 + \,^2C_1 + \,^3C_1 + \,^4C_1 + \,^5C_1$. (b) Express the result as a single combination and compute its numerical value. (c) Verify your answer by adding the five terms individually. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. Row 5: $^5C_0=1$, $^5C_1=5$, $^5C_2=10$, $^5C_3=10$, $^5C_4=5$, $^5C_5=1$ · 2. Sum $= 2^8 = 256$ · 3. $^8C_4 = \frac{8!}{4!4!} = \frac{8\times7\times6\times5}{24} = 70$ · 4. $r=3$, $n=6$: $\sum = \,^7C_4 = 35$ · 5. Each element is in or out of the subset (2 choices each), giving $2^n$ total subsets; $^nC_k$ counts $k$-element subsets, so summing over all $k$ gives all subsets $= 2^n$.
Q1 (1 mark): 1, 5, 10, 10, 5, 1 [1 mark for all six correct].
Q2 (2 marks): (a) $2^9 = 512$ [1]. (b) $^9C_2 = \frac{9 \times 8}{2} = 36$ [1].
Q3 (3 marks): (a) $r = 1$, $n = 5$; hockey-stick gives $\sum_{k=1}^{5}\,^kC_1 = \,^6C_2$ [1]. (b) $^6C_2 = \frac{6 \times 5}{2} = 15$ [1]. (c) $1 + 2 + 3 + 4 + 5 = 15$ ✓ [1]. Note: the hockey-stick with $r = 1$ gives the triangular numbers: $1, 1+2=3, 1+2+3=6, \ldots$
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering Pascal's triangle questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.