IB Maths
Paper 1 & 2
14 min read
Binomial Coefficients & Pascal’s Triangle
Binomial coefficients nCr tell you “how many ways to pick r things from n” ā and they’re the same numbers that pop out when you expand (a + b)n. Pascal’s Triangle is just a visual way to read them off without the formula. Master both ā you’ll need them for the next note.
š
What you need to know
- The formula nCr = n! / [r!(n ā r)!] (in the formula booklet)
- 0! = 1 ā special rule, easy to forget
- nC0 = 1 and nCn = 1 always
- Symmetry: nCr = nCnār (saves you time!)
- Pascal’s Triangle: each row gives the coefficients of (a + b)n
- Your GDC has an nCr button ā use it for big numbers
The binomial coefficient nCr
The formula uses factorials. The factorial symbol ! means “multiply down to 1”:
4! = 4 Ć 3 Ć 2 Ć 1 = 24
And here’s the formula for the binomial coefficient ā let’s break it apart so you can see exactly what each piece does:
nCr
=
n!r! Ć (n ā r)!
1n = total number of items (size of the bigger group).
2r = how many you want to choose. Must satisfy 0 ā¤ r ā¤ n.
3(n ā r)! = the leftovers (the ones you didn’t pick).
!Factorial = product of all positive integers down to 1. 0! = 1.
Computing it by hand ā the cancel trick
Instead of expanding all the factorials, cancel the bigger ones from top and bottom. Watch this calculation of 5C2:
5C2 = 5!2! Ć 3! = 5 Ć 4 Ć 3 Ć 2 Ć 1(2 Ć 1)(3 Ć 2 Ć 1) = 202 = 10
Cancel early! The 3! on the bottom kills the “3 Ć 2 Ć 1” on the top instantly. You’ll save loads of time on non-calc papers.
Properties you must know
These three rules let you skip calculation altogether for many problems.
nC0 = 1
Pick “none” ā only 1 way (because 0! = 1)
nCn = 1
Pick “all” ā only 1 way
nCr = nCnār
Symmetry ā picking “what to take” = picking “what to leave”
Why symmetry helps: instead of computing 10C8 (huge factorials), use 10C8 = 10C2 = 45. Much faster!
The counting principle
nCr answers the question: “How many ways can I choose r objects from n different ones?“
Pick 2 from 5 people
5C2 = 10
10 different pairs are possible.
Pick 3 from 8 students
8C3 = 56
56 different groups of 3.
Pascal’s Triangle
Pascal’s Triangle is a quick visual way to read off binomial coefficients without the formula. Build it with two rules:
- Every row starts and ends with 1.
- Each middle number is the sum of the two numbers directly above it.
Pascal’s Triangle
Row n = the coefficients of (a + b)n. Green: 4 + … = 6 (additive rule). Yellow: row for n = 6.
Top row counts as n = 0, not n = 1! The very top “1” is for (a + b)0 = 1. Many students count rows starting from 1 and end up off by one.
Reading nCr from the triangle
In row n, count from the left starting at position 0. The number you land on is nCr.
For example, in the row 1, 6, 15, 20, 15, 6, 1 (which is n = 6):
- Position 0 ā 6C0 = 1
- Position 1 ā 6C1 = 6
- Position 2 ā 6C2 = 15
- Position 3 ā 6C3 = 20
- Position 4 ā 6C4 = 15
š”
Tips
- For small n (up to 7 or 8), sketching Pascal’s Triangle is faster than the formula
- Use symmetry ā you only need to compute half of each row
- The GDC nCr button handles big numbers instantly. On TI: Math ā PROB ā nCr
- Both the formula and the binomial theorem are in the formula booklet ā don’t waste time memorising
Worked Examples
ā
Example 1 ā Calculate by hand
Without a calculator, find 7C3.
Answer:
Step 1: write the formula with n = 7, r = 3.
7C3 = 7! / [3!(7ā3)!]
= 7! / (3! Ć 4!)
Step 2: cancel the 4! from top and bottom.
= (7 Ć 6 Ć 5 Ć 4!) / (3! Ć 4!)
= (7 Ć 6 Ć 5) / 3!
Step 3: simplify.
= 210 / 6
7C3 = 35
ā
Example 2 ā Use symmetry to save time
Find 10C8 using the symmetry property.
Answer:
Apply symmetry: nCr = nCnār
10C8 = 10C10ā8 = 10C2
Now compute the easier one.
10C2 = 10! / (2! Ć 8!)
= (10 Ć 9) / 2
= 90 / 2
10C8 = 45
Computing 10C8 directly would have meant 8! on the bottom ā way more work.
ā
Example 3 ā Coefficient in a binomial expansion
Without a calculator, find the coefficient of x3 in the expansion of (1 + x)9.
Answer:
Step 1: use the binomial formula with n = 9, a = 1, b = x.
(1 + x)9 = Ī£ 9Cr (1)9ār (x)r
Step 2: xĀ³ occurs when r = 3.
term = 9C3 Ć (1)6 Ć xĀ³
Step 3: compute 9C3 by cancelling.
9C3 = 9! / (3! Ć 6!)
= (9 Ć 8 Ć 7) / (3 Ć 2 Ć 1)
= 504 / 6 = 84
Coefficient of xĀ³ = 84
ā
Example 4 ā Use Pascal’s Triangle
Find the row beginning 1, 6, ā¦ in Pascal’s Triangle and use it to find 6C4.
Answer:
Step 1: build the rows up to n = 6.
n=4: 1, 4, 6, 4, 1
n=5: 1, 5, 10, 10, 5, 1
n=6: 1, 6, 15, 20, 15, 6, 1
Step 2: count from position 0 to position 4 in row n = 6.
Position: 0 ā 1, 1 ā 6, 2 ā 15, 3 ā 20, 4 ā 15
6C4 = 15
Check with symmetry: 6C4 = 6C2 = 15 ā
ā
Example 5 ā Counting principle
A football coach must pick 4 players from a squad of 9 to take penalties. How many different selections are possible?
Answer:
Order doesn’t matter (it’s a selection, not a ranking) ā use nCr.
number of ways = 9C4
Compute by cancelling.
9C4 = 9! / (4! Ć 5!)
= (9 Ć 8 Ć 7 Ć 6) / (4 Ć 3 Ć 2 Ć 1)
= 3024 / 24
126 different selections
ā
Common mistakes
- Forgetting that 0! = 1. This isn’t 0 ā it’s the special rule that makes nC0 work.
- Counting Pascal’s rows from 1. The top row is n = 0, not n = 1. Off-by-one is the #1 mistake here.
- Counting positions from 1. Inside a row, the leftmost number is position r = 0, not r = 1.
- Computing huge factorials by hand. Always cancel ā or use the GDC nCr button.
- Forgetting symmetry. 20C18 looks scary but is the same as 20C2 = 190.
- Mixing up nCr with nPr. nCr ignores order (combinations); nPr cares about order (permutations).
Final word: the formula is in the booklet, the GDC has the button, and Pascal’s Triangle is fast for small n. Pick the easiest method for the question ā examiners won’t reward extra factorial gymnastics. Next note: how these coefficients power the full binomial theorem.
Need help with the Binomial Theorem?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
