At Maryland Middle School, there are 9 mathematics classes. If two mathematics classes are selected for a school competition, what is the number of possible outcomes?
combinations of 2 out of 9
C(n,r) = n!/[(n-r)! (r!) ]
= 9! /[ 7! * 2! ]
= 9*8 *7! / [ 7! * 2]
=9*4
= 36
Or use Pascal's triangle
row 9 is
1 9 36 84 126 126 84 36 9 1
so r = 0 --> 1
r = 1 = 9 of course
r = 2 = 36
To find the number of possible outcomes when selecting two mathematics classes out of 9, we can use the combination formula. The formula for calculating combinations is given by:
C(n, r) = n! / (r! * (n-r)!)
Where:
n is the total number of classes (9 in this case),
r is the number of classes to be selected (2 in this case), and
! denotes factorial.
Plugging in the values, we get:
C(9, 2) = 9! / (2! * (9-2)!)
Simplifying further:
C(9, 2) = 9! / (2! * 7!)
The factorial of a number is the product of all positive integers less than or equal to that number. So:
9! = 9 * 8 * 7!
Now we substitute this into the equation:
C(9, 2) = (9 * 8 * 7!) / (2! * 7!)
The 7! term cancels out:
C(9, 2) = (9 * 8) / (2 * 1)
Evaluating the expression:
C(9, 2) = 72 / 2
C(9, 2) = 36
Therefore, there are 36 possible outcomes when selecting two mathematics classes from the 9 available at Maryland Middle School.
To determine the number of possible outcomes, we can use the concept of combinations. In this case, we need to select two mathematics classes out of the nine available classes for the competition.
The formula to calculate the number of combinations is:
\[C(n, r) = \frac{{n!}}{{r!(n-r)!}}\]
Where:
- n represents the total number of items (mathematics classes)
- r represents the number of items to be selected (classes for the competition)
- ! denotes factorial, which means multiplying a number by all positive integers less than itself down to 1
Applying this formula to our situation, we have:
\[C(9, 2) = \frac{{9!}}{{2!(9-2)!}}\]
Simplifying further:
\[C(9, 2) = \frac{{9!}}{{2!7!}}\]
Now, let's calculate the factorials:
\[9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1\]
\[2! = 2 × 1\]
\[7! = 7 × 6 × 5 × 4 × 3 × 2 × 1\]
Plugging these values into the formula:
\[C(9, 2) = \frac{{9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1}}{{(2 × 1)(7 × 6 × 5 × 4 × 3 × 2 × 1)}}\]
Canceling out similar terms:
\[C(9, 2) = \frac{{9 × 8}}{{2 × 1}}\]
Finally, calculating:
\[C(9, 2) = 36\]
Therefore, there are 36 possible outcomes when selecting two mathematics classes for the school competition at Maryland Middle School.