You need to know these two formulas:
![\huge a^b \times a^c = a^{b+c}\\\\\huge \frac{a^b}{a^c}=a^{b-c}](https://tex.z-dn.net/?f=%5Chuge%20a%5Eb%20%5Ctimes%20a%5Ec%20%3D%20a%5E%7Bb%2Bc%7D%5C%5C%5C%5C%5Chuge%20%5Cfrac%7Ba%5Eb%7D%7Ba%5Ec%7D%3Da%5E%7Bb-c%7D)
For example,
![5^2 \times 5^3 = 5^{2+3} = 5^5](https://tex.z-dn.net/?f=5%5E2%20%5Ctimes%205%5E3%20%3D%205%5E%7B2%2B3%7D%20%3D%205%5E5)
Answer:
40951
Step-by-step explanation:
Using the principles of inclusion - Exclusion
Where C(n,r)=n!/(n-r)!r!
Total elements in the five sets including number repetition is given as (10000)×C(5, 1) =10000× 5!/(5-1)!1!=10000×5=50000
Total Number of elements in each pair including number repetition of sets is given as
=(1000) × C(5, 2) =10000
Number of elements in each triple of sets is given as
=(100) × C(5, 3) =1000
Number of elements in every four sets
=(10) × C(5, 4)=50
Number of elements in every one set
(1) × C(5, 5)=1
Therefore total number of unique elements=50000-10000+1000-50+1
=40951
Answer: a) An = An-1 + An-2
b) 55ways
Step-by-step explanation:
a) a nickel is 5 cents and a dime is 10cent so a multiple of 5 cents is the possible way to pay the tolls in both choices.
Let An represents the number of possible ways the driver can pay a toll of 5n cents, so that
An = 5n cents
Case 1: Using a nickel for payment which is 5 cents, the number of ways given as;
An-1 = 5( n-1)
Case 2: using a dime which is two 5 cents, the number of ways is given as;
An-2 = 5(n-2)
Summing up the number of ways, we have
An = An-1 + An-2
From the relation,
If n= 0, Ao= 1
n =1, A1= 1
b) 45 cents paid in multiples of 5cents will give us 9 ways(A9)
From the relation, we have that
Ao = 1
A1 = 1
An =An-1 + An-2
Ao = 1
A1 = 1
A2 = A1+Ao = 1+1= 2
A3 = A2 + A1 = 3
A4 = A3+A2=5
A5=A4+A3=8
A6=A5+A4=13
A7 =A6+A5 = 21
A8= A7+A6= 34
A9= A8+A7= 55
So there are 55ways to pay 45cents.
20x4=80
i am assuming this is correct this is how i learned just take all numbers in equation and multiply by each other