Worked examples
The question
Question: work out 5 C 2.
5C2 =
1 Write the rule
The rule for combinations: n factorial, over r factorial times n minus r, factorial.
nCr = n!r! × (n − r)!
Here n is 5 and r is 2.
n = 5
r = 2
2 Put the numbers in
We put 5 for n and 2 for r.
5C2 = 5!2! × (5 − 2)!
5 minus 2 is 3.
5C2 = 5!2! × 3!
3 Write out and cancel
We write every factorial out.
5C2 = 5 × 4 × 3 × 2 × 1(2 × 1) × (3 × 2 × 1)
3 times 2 times 1 is on top and at the bottom.
They cancel.
What is left is 5 × 4, over 2 × 1.
5C2 = 5 × 42 × 1
Why divide by 2 factorial?
The top, 5 times 4, counts in order.
One group of 2 can be put in order 2 ways.
So each group is counted 2 times.
one group of 2: 2 × 1 = 2 orders
AB, BA
Careful!
5 times 4 counts each group 2 times.
We must divide by 2.
Careful: not 5 × 4 = 20 ✗
4 Work it out
2 times 1 is 2.
4 on top and 2 at the bottom both divide by 2.
What is left is 5 × 2.
5C2 = 5 × 2
5 times 2 is 10.
5 × 2 = 10
So 5 C 2 is 10.
5C2 = 10
5 Check
Let's check: 10 groups, each in 2 orders, make 20.
10 groups × 2 orders = 20
And 5 times 4, in order, is 20.
The same.
So the answer is right!
5 × 4 = 20 ✓
Let's look back.
First: write the rule.
Then: put the numbers in.
Then: write out and cancel.
Then: work it out.
Last: check.
▶ Watch it solved
The question
Question: work out 8 C 3.
8C3 =
1 Write the rule
The rule for combinations: n factorial, over r factorial times n minus r, factorial.
nCr = n!r! × (n − r)!
Here n is 8 and r is 3.
n = 8
r = 3
2 Put the numbers in
We put 8 for n and 3 for r.
8C3 = 8!3! × (8 − 3)!
8 minus 3 is 5.
8C3 = 8!3! × 5!
3 Write out and cancel
We write every factorial out.
8C3 = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1(3 × 2 × 1) × (5 × 4 × 3 × 2 × 1)
5 times 4 times 3 times 2 times 1 is on top and at the bottom.
They cancel.
What is left is 8 × 7 × 6, over 3 × 2 × 1.
8C3 = 8 × 7 × 63 × 2 × 1
Why divide by 3 factorial?
The top, 8 times 7 times 6, counts in order.
One group of 3 can be put in order 6 ways.
So each group is counted 6 times.
one group of 3: 3 × 2 × 1 = 6 orders
ABC ACB BAC
BCA CAB CBA
Careful!
8 times 7 times 6 counts each group 6 times.
We must divide by 6.
Careful: not 8 × 7 × 6 = 336 ✗
4 Work it out
3 times 2 times 1 is 6.
6 on top and 6 at the bottom both divide by 6.
What is left is 8 × 7.
8C3 = 8 × 7
8 times 7 is 56.
8 × 7 = 56
So 8 C 3 is 56.
8C3 = 56
5 Check
Let's check: 56 groups, each in 6 orders, make 336.
56 groups × 6 orders = 336
And 8 times 7 times 6, in order, is 336.
The same.
So the answer is right!
8 × 7 × 6 = 336 ✓
Let's look back.
First: write the rule.
Then: put the numbers in.
Then: write out and cancel.
Then: work it out.
Last: check.
▶ Watch it solved
The question
Question: work out 9 C 3.
9C3 =
1 Write the rule
The rule for combinations: n factorial, over r factorial times n minus r, factorial.
nCr = n!r! × (n − r)!
Here n is 9 and r is 3.
n = 9
r = 3
2 Put the numbers in
We put 9 for n and 3 for r.
9C3 = 9!3! × (9 − 3)!
9 minus 3 is 6.
9C3 = 9!3! × 6!
3 Write out and cancel
We write every factorial out.
9C3 = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1(3 × 2 × 1) × (6 × 5 × 4 × 3 × 2 × 1)
6 times 5 times 4 times 3 times 2 times 1 is on top and at the bottom.
They cancel.
What is left is 9 × 8 × 7, over 3 × 2 × 1.
9C3 = 9 × 8 × 73 × 2 × 1
Why divide by 3 factorial?
The top, 9 times 8 times 7, counts in order.
One group of 3 can be put in order 6 ways.
So each group is counted 6 times.
one group of 3: 3 × 2 × 1 = 6 orders
ABC ACB BAC
BCA CAB CBA
Careful!
9 times 8 times 7 counts each group 6 times.
We must divide by 6.
Careful: not 9 × 8 × 7 = 504 ✗
4 Work it out
9 times 8 times 7 is 504.
3 times 2 times 1 is 6.
9C3 = 5046
9 times 8 is 72.
72 times 7 is 504.
9 × 8 = 72
72 × 7 = 504
504 divided by 6 is 84.
504 ÷ 6 = 84
So 9 C 3 is 84.
9C3 = 84
5 Check
Let's check: 84 groups, each in 6 orders, make 504.
84 groups × 6 orders = 504
And 9 times 8 times 7, in order, is 504.
The same.
So the answer is right!
9 × 8 × 7 = 504 ✓
Let's look back.
First: write the rule.
Then: put the numbers in.
Then: write out and cancel.
Then: work it out.
Last: check.
▶ Watch it solved
