nC0 = 1 where n is any positive integer. This means that there is one way of choosing no objects from a set of objects of any size, for example, 5C0 = 5!/5!(5-5)! = 120/120 = 1.
nCn = 1. There is one way of choosing all the objects in a set of any size. That is, there is only one way to choose all the items, which is to choose all the items.
No other binomial coefficient equals 1, for example, 4C4 = 4!/(4-4)!0! = 24/24 = 1.
nCk = nC(n-k), for example, 5C3 = 5!/3!(5-3)! = 120/6*2 = 10. Here n = 5 and k = 3. Substituting n-k for k gives 5C2, which equals 5!/(5-3)!2! = 120/2*6 = 10. This makes sense. Suppose you have 5 friends and you choose 3 to invite to dinner. You could just as easily have chosen which 2 to not have invited to dinner.
One way to form Pacal's triangle is to start with a 1 in the top row, two 1s in the next row, and, for each succeeding row, adding the number immediately above and the number above and to the left to get a new number. The first few rows are (see Resource for proper formatting of the triangle):
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
The rows of the triangle are the binomial coefficients. For example, the last row shown above gives 4C0 = 1, 4C1 = 4, 4C2 = 6, 4C3 = 4 and 4C4 = 1.
Pascal's triangle, named after Blaise Pascal, a French mathematician, has many interesting properties. One is that, if you add up the numbers in each row, you get the Fibonacci sequence.
nC0 + nC1 + nC2 + .... nCn = 2^n, for example, 4C0 + 4C1 + 4C2 + 4C3 + 4C4 = 1 + 4 + 6 + 4 + 1 = 16 = 2^4. One way of thinking of this is to imagine that you are "choosing" a number of heads from a number of coin tosses, so that 4C0 would mean "no heads in four tosses." There are two ways a coin can fall on each toss, so there are 2^n ways that a coin can fall on n tosses. Further, you can get any number of heads, from no heads to all heads, so the sum of the binomial coefficients equals 2^n.