The triangle in one picture
Pascal’s triangle starts with a single 1 at the top. Every other entry is the sum of the two entries directly above it, and the edges are always 1. The first seven rows look like this:
| Row | Entries |
|---|---|
| 0 | 1 |
| 1 | 1, 1 |
| 2 | 1, 2, 1 |
| 3 | 1, 3, 3, 1 |
| 4 | 1, 4, 6, 4, 1 |
| 5 | 1, 5, 10, 10, 5, 1 |
| 6 | 1, 6, 15, 20, 15, 6, 1 |
Row 5, for example, is built from row 4: 1, then 1 + 4 = 5, then 4 + 6 = 10, then 6 + 4 = 10, then 4 + 1 = 5, then 1.
What the entries mean
The entry in row n at position k (counting positions from 0) tells you how many ways there are to choose k items from n. It is written C(n, k), or “n choose k”, and these numbers are the binomial coefficients. So in row 5 the entry at position 2 is 10: there are 10 ways to choose 2 items from 5.
Here is the link to repeated trials. Flip a coin 5 times. Each flip is heads or tails, so there are 2 × 2 × 2 × 2 × 2 = 32 equally likely sequences. To get exactly 2 heads, you must pick which 2 of the 5 flips are heads. That is a choice of 2 positions from 5, and there are C(5, 2) = 10 ways to do it. So the chance of exactly two heads is 10/32 = 0.3125, or 31.25%.
Reading across row 5 gives the full picture for five flips:
- 0 heads: 1 way
- 1 head: 5 ways
- 2 heads: 10 ways
- 3 heads: 10 ways
- 4 heads: 5 ways
- 5 heads: 1 way
The total is 1 + 5 + 10 + 10 + 5 + 1 = 32, which matches 2^5. That is not a coincidence: every row n adds up to 2^n.
Why the building rule works
The rule C(n, k) = C(n − 1, k − 1) + C(n − 1, k) has a plain explanation. Imagine choosing k players from n, and single out one particular player. Either that player is on the team, leaving k − 1 to pick from the other n − 1, or they are not, leaving k to pick from the other n − 1. Those two cases cover everything and do not overlap, so you add them.
Check it with row 5: C(5, 2) = C(4, 1) + C(4, 2) = 4 + 6 = 10, which agrees with the table.
Symmetry and the middle
Each row reads the same forwards and backwards, because choosing k items to include is the same as choosing the n − k items to leave out. In row 6, C(6, 2) = 15 and C(6, 4) = 15.
The entries are largest near the middle. For six flips of a fair coin, there are 20 sequences with exactly three heads out of 64, so the chance is 20/64 = 0.3125. Only 1 sequence in 64 has no heads at all. This is why results near an even split are the most common, though any individual exact count can still be fairly unlikely.
Computing an entry without the whole triangle
For large rows, building the triangle is slow. The standard formula uses factorial notation:
C(n, k) = n! ÷ (k! × (n − k)!)
For C(5, 2): 5! = 120, 2! = 2 and 3! = 6, so the result is 120 ÷ 12 = 10. A shortcut avoids the large factorials by cancelling: C(7, 3) = (7 × 6 × 5) ÷ (3 × 2 × 1) = 210 ÷ 6 = 35.
Where players meet these numbers
Choosing which items from a group are selected is a combination question, so these coefficients appear whenever order does not matter. Examples include the number of different two-card starting hands, which is C(52, 2) = 1,326, and the number of ways to pick a set of numbers in a lottery-style draw. Counting how many hand combos are consistent with a given range is the same kind of calculation.
They also feed straight into the binomial distribution, where each coefficient is multiplied by the probability of one specific sequence. And for tasks like checking whether a set of results is surprising, the triangle gives a quick sense of how many sequences produce a given count.
Pascal’s triangle and fairness
The triangle assumes each trial is independent and has the same chance. A fair coin flipped five times fits that model. Rolling a die and asking for “a six or not a six” does not give equal chances to the two outcomes, so the number of sequences is no longer a direct probability; each sequence needs its own weight.
Also remember that earlier results do not change the odds of the next trial. After four heads in a row, a fair coin is still 50/50 on the fifth flip.