You have seven kinds of chocolate and you want to know how many combinations of three kinds can you make. Binomial coefficients are the mathematical way of picking k-item subsets from n original items. The common symbol for a binomial coefficient looks like a fraction inside parentheses without the fraction bar (n k), or nCk, and is read "n choose k." N choose k gives the number of k-subsets that are possible out of n different objects or choices. Use the following formula: nCk = n!/[(n-k)!*k!].

- Skill level:
- Moderate

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## Instructions

- 1
Set up an example, such as 5C3. Write 5C3 out explicitly in factorial form.

From the problem, you have n = 5 and k = 3.

The factorial form is 5!/[(5-3)!*3!].

- 2
Solve what's in the parentheses first: (5-3) = 2, so the problem now reads: 5!/[2!*3!].

- 3
Calculate n! (that is, n factorial), where n! = n

*(n-1)*(n-2)*...*2*1.In the example problem: n=5, so n! = 5! = 5

*4*3*2*1 = 20*3*2*1 = 60*2*1 = 120*1 = 120. Now the problem reads: 120/[2!*3!]. - 4
Calculate (n-k)! In the example problem, (n-k)! = 2! from Step 2.

2! = 2

*1 = 2. Now the problem reads: 120/[2*3!]. - 5
Calculate k! In the example problem, k! = 3! = 3

*2*1 = 6*1 = 6.Now the problem reads: 120/[2*6].

- 6
Solve the problem inside the brackets. In the example problem, [2*6] = 12

Now the problem reads: 120/12.

- 7
Simplify the fraction using division. In the example problem, 120/12 = 10

#### Tips and warnings

- To solve binomial coefficients, you must be familiar with factorials. Do not attempt this without prior knowledge of factorials. For example: 5! means "5 factorial" and 5! = 5*4*3*2*1. If x is any positive integer, then x! = x*(x-1)*(x-2)*...*3*2*1.