Jen,amy,linda,max and jenny are compteting to fill two remaining spots on the chess team. How mant different combinations are possible for filling the two spots?
120,60,20,or 10
Let s = spots and n = number of students
n! = "n factorial" = product of all numbers n or less
P = n!/s!(n-s)! = 5!/2!(5-2)!
5! = 5 * 4 * 3 * 2 *1 = 120
2! = 2 * 1 = 2
(5-2)! = 3! = 3 * 2 * 1 = 6
P = 120/12 = 10
To find the number of different combinations for filling the two spots on the chess team, we can use the concept of combinations.
In this scenario, we have 5 individuals (Jen, Amy, Linda, Max, and Jenny) competing for 2 spots. To calculate the number of combinations, we can use the formula for combinations, which is:
C(n, r) = n! / (r! * (n-r)!)
Where:
- n is the total number of individuals (5 in this case)
- r is the number of spots available (2 in this case)
- ! represents the factorial of a number (i.e., the product of all positive integers up to that number)
Substituting the values into the formula, we get:
C(5, 2) = 5! / (2! * (5-2)!)
= 5! / (2! * 3!)
Now, let's find the factorial values:
5! = 5 * 4 * 3 * 2 * 1 = 120
2! = 2 * 1 = 2
3! = 3 * 2 * 1 = 6
Plugging in these values, we have:
C(5, 2) = 120 / (2 * 6)
= 120 / 12
= 10
Therefore, there are 10 different combinations possible for filling the two spots on the chess team.