If we start with 10 choices, have 5 decisions to make/positions to fill, and repetition isn’t allowed, how many combinations can be made?
252?
yes
and with repetition, it is 2002?
how did you get 2002 ?
You are making decisions and repetition is allowed?
Are you making the same decisions more than once??
it asks this:
How many combinations are possible with 10 initial choices and 5 decisions to make if repetition is allowed?
i used
(r + n – 1)!/r!(n – 1)!
(5 + 10 - 1)!/5!(10 - 1)!
To determine the number of combinations that can be made with 10 choices, 5 decisions to make, and no repetition allowed, we can use the formula for combinations.
The formula for combinations is given by:
C(n, r) = n! / (r! * (n - r)!)
Where:
C(n, r) represents the number of combinations
n is the total number of choices
r is the number of decisions or positions to fill
! represents the factorial operation
Using the given information, we have:
n = 10 (10 choices)
r = 5 (5 decisions to make)
Plugging these values into the formula, we can calculate the number of combinations:
C(10, 5) = 10! / (5! * (10 - 5)!)
= 10! / (5! * 5!)
To calculate the factorial, we multiply all the numbers from the given value down to 1.
10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
5! = 5 * 4 * 3 * 2 * 1
Substituting the factorial values into the equation:
C(10, 5) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / (5 * 4 * 3 * 2 * 1 * 5 * 4 * 3 * 2 * 1)
= 10 * 9 * 8 * 7 * 6 / (5 * 4 * 3 * 2 * 1)
= 3024 / 120
= 252
Therefore, there are 252 combinations that can be made with 10 choices, 5 decisions to make, and no repetition allowed.