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!].
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Things you need
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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!].
Solve what's in the parentheses first: (5-3) = 2, so the problem now reads: 5!/[2!*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! = 54321 = 20321 = 6021 = 1201 = 120. Now the problem reads: 120/[2!*3!].
Calculate (n-k)! In the example problem, (n-k)! = 2! from Step 2.
2! = 21 = 2. Now the problem reads: 120/[23!].
Calculate k! In the example problem, k! = 3! = 321 = 6*1 = 6.
Now the problem reads: 120/[2*6].
Solve the problem inside the brackets. In the example problem, [2*6] = 12
Now the problem reads: 120/12.
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.
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