We're going to be using combination since this question is asking how many different combinations of 10 people can be selected from a set of 23.
We would only use permutation if the order of the people in the committee mattered, which it seems it doesn't.
Formula for combination:
Where 
represents the number of objects/people in the set and 
represents the number of objects/people being chosen from the set
There are 23 people in the set and 10 people being chosen from the set
Usually I would prefer solving such fractions by hand instead of a calculator, but factorials can result in large numbers and there is too much multiplication. Using a calculator, we get
Thus, there are 1,144,066 different 10 person committees that can be selected from a pool of 23 people. Let me know if you need any clarifications, thanks!
~ Padoru
Answer:
Step-by-step explanation:
look this up it give you the answer
let me know if this help plzz
2.25 x 1000 = 2250.
That's how much yds of ribbons he needs.
An ordered pair (x,y) is a solution to a system of equations if it makes all the equations true.
Let's check whether (–1, 5) makes the equations true.
Plugging –1 in the first equation for x and 5 in for y, we get
–1 + 5 = 4: TRUE
Plugging –1 in the second equation for x and 5 in for y, we get
–1 – 5 = –6: TRUE
Since it makes both the equations true, it's a solution to the system of equations. So the answer choice is D, the 4th one.
Find the zeros<span> of Function (</span>x)= (x+3)^2(x-5)^6 and state the multiplicity. ... (x--5)^6<span> = 0=> </span>x--5<span>= 0 => </span>x<span> = </span>5<span>. Therefore real </span>zero's of the function<span> are --</span>3<span>,5.</span><span />
