A cab driver is paid $6 plus $0. 45 per mile driven. Which expression shows the amount of money the cab driver earns in m miles?
1 answer:
Algebraic expression is the expression which consist the variables, coefficients and the constants. The expression for the condition given in the problem is .
Given information
A cab driver is paid $6 plus $0.45 per mile driven.
Total number of total miles is <em>m.</em>
<h3>Algebraic expression</h3>Algebraic expression is the expression which consist the variables, coefficients and the constants. The algebraic expression are used to solve the condition based problems by expressing them into the algebraic form.
As the cab driver is paid $6 as the fixed amount of the taxi irrespective to the mile driven. Thus this is independent to the other variables and written as the constant in the expression.
As the total miles driven by the cab driver is<em> m</em>. The cost of the per mile driven the car is $0.45. Thus the expression for this condition can be given as,
Hence, the expression for the condition given in the problem is .
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a. Your answer is correct, 26 choices for each of 13 positions, so total possible strings.
b. You have the right idea, but your method only counts one type of permutation, like
C H A R I T Y _ _ _ _ _ _
but doesn't account for other arrangements like
_ _ _ C H A R I T Y _ _ _
or
_ _ _ _ _ C H A R I T Y _
Treating CHARITY as one letter, we're then considering strings of length 7 (6 open slots plus this string), which we can arrange in 7! different ways. So the total number of such strings is .
c. This one is a bit more involved. I would go about it by counting the number of strings containing CHARITY but not HORSES, HORSES but not CHARITY, and both CHARITY and HORSES.
- CHARITY but not HORSES
As we know from part (a), there are strings containing CHARITY, but the string HORSES can be found whenever there are 6 open slots to either side of CHARITY, i.e. in strings of either form
C H A R I T Y _ _ _ _ _ _
or
_ _ _ _ _ _ C H A R I T Y
Then there are 2 strings that we want to remove from the count, giving such strings.
- HORSES but not CHARITY
Reasoning as we did in part (b) suggests that there are possible strings containing HORSES, and reasoning as we did in the previous case suggests only 2 of these contain CHARITY, giving a total of
such strings.
- CHARITY and HORSES
There are 2 such strings,
C H A R I T Y H O R S E S
H O R S E S C H A R I T Y
Then by the inclusion-exclusion principle, the number of strings containing either CHARITY or HORSES is .
Finally, the number of strings containing neither CHARITY nor HORSES is complementary to the number of strings containing either of them, so the total is .