Ms. Washington has 18 students in her class. She wants to send 3 of her students to pick up books for the class. How many combinations of students can she choose?
816
54
306
4896
I'm not sure if its A?
816 is correct, but I am interested to know why you decided on A.
ty.
18c3=18!/3!15! =816
To find the number of combinations of students Ms. Washington can choose, we can use the combination formula.
The combination formula is given by:
nCr = n! / (r!(n-r)!)
Where n is the total number of students (18 in this case), and r is the number of students that Ms. Washington wants to send to pick up books (3 in this case).
Now let's substitute the values into the formula and calculate:
nCr = 18! / (3!(18-3)!)
= 18! / (3! * 15!)
To simplify further, we can use the fact that 15! can be canceled out from both the numerator and denominator:
nCr = (18 * 17 * 16 * 15!) / (3! * 15!)
= (18 * 17 * 16) / 3!
The value of 3! (3 factorial) is 3 * 2 * 1 = 6, so we can substitute that into the equation:
nCr = (18 * 17 * 16) / 6
= 4896 / 6
= 816
Therefore, Ms. Washington can choose 816 different combinations of students to send to pick up books.
So your answer is indeed A, 816.