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:
a. 13/3
b. 3 3/4
d. 1 7/8
c. 7/2
Step-by-step explanation:
a. 4 1/3 given
4*3=12 multiply the whole numbers with the denominator
12+1 =13 add the numerator to the product
13/3 put sum over denominator
b. 15/4 given
15/4=3 with a remainder of 3 divide numerator by denominator
3 3/4 rewrite as mixed number
d. 15/8 given
15/8=1 with a remainder of 7 divide numerator by denominator
1 7/8 rewrite as mixed number
c. 3 1/2 given
3*2=6 multiply whole number by denominator
6+1=7 add numerator to product
7/2 put sum over denominator
Answer:
p=5
Step-by-step explanation:
a(x+y) is also (x+y)a
So if 3(n+5) has to be equivalent to (n+p)3, then p has to be 5.
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hope it helps
Answer:
yes to 5
Step-by-step explanation:
divide 45 by 9 and u get 5
