Math Combinations- Formula and Example Problemsدوره: GRE Test- Practice & Study Guide / فصل: GRE Quantitative Reasoning- Probability and Statistics / درس 9
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.
- زمان مطالعه 7 دقیقه
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متن انگلیسی درس
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.
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.
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
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.
With such a tough season ahead of them, the Wildcats coach knows that his team must have a lot of practice. He decided that the team would play three-on-three games to work on their skills. There are 12 players on the team, and three of them will be chosen for each team. The coach now needs to know how many combinations of teams he could create. To use the equations, the variable n would equal 12 and the r variable would equal 3.
The coach needs to use the equation nCr = n!/r!(n-r)! . The coach then will need to substitute 12 in for n and 3 in for r . Next, he will need to subtract 12-3 = 9. So, he now has 12!/(3! x 9!).
Now, let’s expand the factorials. 12! = 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 and 9! x 3! = 3 x 2 x 1 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.
The easiest way to calculate the combination is to cancel out common terms. Since 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 appears on both the top and bottom of the equation, those terms can be cancelled out. The coach now has 12 x 11 x 10 on top and 3 x 2 x 1 left on bottom. The coach now multiplies on top, which equals 1,320, and the bottom, which equals 6. Next, the coach divided 1,320 / 6 = 220. The coach could choose 220 different combinations of three-player teams.
Just to review, a combination is an arrangement of objects where order does not matter. To calculate a combination, you must use a factorial. A factorial is the product of all the positive integers equal to and less than your number. The equation to calculate the 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.
When solving the equation, remember that the easiest way to solve is by canceling out common terms on both the top and the bottom of your equation.
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