Math Combinations: Formula and Example Problems
Combinations are an arrangement of objects where order does not matter. In this lesson, the coach of the Wildcats basketball team uses combinations to help his team prepare for the upcoming season.
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Basketball Season and Combinations
The Jackson Wildcats play basketball in a highly competitive city district. There are eight teams in the district, and they all play each other once during the season. The coach of the Wildcats wants to know how many games will be played in the district this season. To calculate this amount, he will need to use a combination.
Combinations
A combination is an arrangement of objects where order does not matter. The coach knows that there are eight teams, but the order the teams play each other does not matter. One way the coach could calculate the number of games is to list out each team and the teams they would play. The coach realized, though, that there were some games that would be repeated when writing them out. So, he researched and found a formula to calculate the number of combinations. The formula for a combination is nCr = n!/r!(n-r)!, where n represents the number of items and r represents the number of items being chosen at a time.
Factorial
To calculate a combination, you must know how to calculate a factorial. A factorial is the product of all the positive integers equal to and less than your number. A factorial is written as the number followed by an exclamation point. For example, to write the factorial of 6, you would write 6!. To calculate the factorial of 6, you would multiply all of the positive integers equal to and less than 6.
6! = 6 x 5 x 4 x 3 x 2 x 1
By multiplying these numbers together, we find that 6! = 720. Let's look at another example of how we would write and solve the factorial of 11. The factorial of 11 would be written as 11!. To calculate:
11! = 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 39,916,800
Combination Formula
The coach of the Wildcats now knows that he has to use the equation nCr = n!/r!(n-r)!, where n represents the number of items and r represents the number of items being chosen at a time. Using this equation, he must select two teams for each game from the eight teams in the district. So, the variable n would equal 8 and the variable r would equal 2. The equation would then look like 8 C 2 = 8!/2!(8-2)!.
To solve this equation, we would first need to perform (8-2) in the parenthesis, which would equal (8!/(2! x 6!). Next, we would expand 8!, 2! and 6!. 8! would equal 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, over 2!, which is 2 x 1 x 6 x 5 x 4 x 3 x 2 x 1. By multiplying 8! on the top, it equals 40,320, and 2! x 6! on the bottom equals 1,440. Finally, we would divide 40,320 by 1,440, which would equal 28. The Wildcats coach now knows that there are 28 games that will be played in their district this season.