nCr · Counting & Combinatorics
Counting every possibility — without omission or repetition.
3 shirts and 2 pairs of pants — how many outfits? 3+2=5? 3×2=6?
8,145,060 — the denominator behind a lottery jackpot.
Open the immersive version →Why does this question matter?
Probability is 'favorable cases ÷ total cases' — numerator and denominator are both counts. Miscount and the whole probability collapses. The letters between Pascal and Fermat answering gamblers' questions founded this field.
But counting by listing hits a wall fast. You cannot list 8.14 million lottery combinations — laws for counting without listing were needed.
Three tools suffice for astronomically large counts: the product rule, the sum rule, and the question of order (permutations vs combinations). Password strength, gene combinations, network routes — all counted with these.
Pause and try it yourself
Make a prediction first. Change one input at a time and compare what happens with what you expected.
Predict first — 3 shirts, 2 pairs of pants. How many outfits?
Fill in the outfits one by one
Tap the grid cells to complete each outfit. Let's count how many get filled.
outfits filled: 0 / 6
Read the idea from another angle
Counting is math's oldest skill and its easiest to get wrong.
Miss a case (omission) or count one twice (repetition) — counting has exactly these two failure modes. The product/sum rules and permutations/combinations are the safety devices against both.
The lottery's 8.14 million comes from calculation, not listing.
45×44×43×42×41×40 divided by 6! is 8,145,060. One line of arithmetic captures a cosmic number of cases — the power of counting without listing.
Now read the definitions with their conditions
The rules discovered in the outfit grid and the picking experiments, formalized.
Product rule · sum rule
Pass through every stage: multiply. Pick one branch: add. The two pillars of all counting.
Permutations (order matters)
Choose r from n and line them up: n ways for the first seat AND (n−1) for the next… — a chain of the product rule.
Combinations (order ignored)
Ignoring order, each group was counted r! times — so divide by r!. The identity of the lottery's 8.14 million.
Make one more distinction
Multiple stages of choices always add.
'And' (passing through every stage) multiplies; 'or' (picking one branch) adds. Wear a shirt AND pants: 3×2. Choose chicken OR pizza: 3+2.
Choosing 2 from 3 people is always 3×2=6.
For president & vice-president (order matters), 6 is right. For two cleaning-duty members (order doesn't), (Amy,Ben) and (Ben,Amy) are the same case — 6÷2=3. Asking 'does order matter?' is half of counting.
Where else can this idea help?
The denominator of probability
Lottery jackpot 1/8,145,060, royal flush 1/649,740 — every probability's denominator is a count. Wrong count, wrong probability.
Password strength
4 digits: 10⁴ = 10,000 cases. 8 alphanumerics: 36⁸ ≈ 2.8 trillion. The time to try them all IS the security — counting is the shield.
Genetics
Parental gene combinations, DNA sequences (4ⁿ) — genetics has run on counting since Mendel's peas.
Scheduling and routing
Visiting 10 cities has 10! = 3.6 million orderings. Counts explode combinatorially — which is why optimization algorithms exist. Combinatorial explosion is central to computer science.
Explain it in your own words
Try another question
Q14 mains and 3 drinks — how many set menus?Show answer ▾
Pick a main AND a drink: 4×3 = 12. If you picked only a main OR only a drink, it would be 4+3 = 7.
Q2From 5 people, how many ways to pick a president and vice-president? And to pick 2 delegates?Show answer ▾
President & VP (order matters): 5×4 = 20. Two delegates (order doesn't): 20÷2 = 10. The same 'pick 2' changes answer with the order question.
Q3Flipping a coin 10 times, how many possible outcome sequences are there?Show answer ▾
2 choices chained 10 times by 'and': 2¹⁰ = 1,024. The product rule gives birth to powers — counting meets exponents.
Try answering yourself before revealing it — getting it wrong is where learning starts.
What question comes next?
Leads to next
Probability
You've learned to count; now divide — favorable ÷ total, putting a number on uncertainty.
Go to the Probability lab →Related lab
Probability
Probability = favorable ÷ total — counting supplies the denominator.
Go to the Probability lab →Related lab