If we start with 10 choices, have 5 decisions to make/positions to fill, and repetition isn’t allowed, how many combinations can we make?
By the definition of combinations that would be "10 choose 5"
= C(10,5) or 10!/(5!5!) = ...
To determine the number of combinations possible when making multiple choices without repetition, you need to use the concept of permutations.
In this case, you have 10 choices and need to make 5 decisions. Since repetition is not allowed, each choice can only be used once.
To calculate the number of combinations, you can use the formula for permutations without repetition, which is denoted as P(n, r), where "n" represents the total number of choices and "r" represents the number of decisions/positions to be filled.
The formula for permutations without repetition is:
P(n, r) = n! / (n - r)!
where "!" denotes the factorial operation.
In this scenario, you want to calculate P(10, 5).
Substituting the values into the formula:
P(10, 5) = 10! / (10 - 5)!
We can expand the factorials:
P(10, 5) = 10! / 5!
P(10, 5) = (10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (5 x 4 x 3 x 2 x 1)
After canceling out the common factors:
P(10, 5) = (10 x 9 x 8 x 7 x 6) / (5 x 4 x 3 x 2 x 1)
Calculating the values:
P(10, 5) = 30240 / 120
Simplifying the division:
P(10, 5) = 252
Therefore, you can make 252 different combinations when selecting 5 choices out of 10 without repetition.