9_12-15-22-5_25-15-21translate the code
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Answer:
Step-by-step explanation:
Lets divide it in cases, then sum everything
Case (1): All 5 numbers are different
In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.
The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.
Case (2): 4 numbers are different
We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4
We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.
The total cardinality of this case is
Case (3): 3 numbers are different
As we did before, we pick 3 elements from a set of n. The amount of possibilities is
Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of
Case (4): 2 numbers are different
We pick 2 numbers from a set of n, with a total of possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.
The total amount of possibilities for this case is
Case (5): All numbers are the same
This is easy, he have as many possibilities as numbers the set has. In other words, n
Conclussion
By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is
I hope that works for you!
Marilyn's finance charge at the end of the first month will be
$991.38 × 0.199/12 = $16.44
The balance subject to the next month's finance charge will be
$991.38 +16.44 -410.00 = $597.82
The finance charge at the end of the second month will be
$597.82 × 0.199/12 = $9.91
The balance remaining after the second payment will be
$597.82 +9.91 -410.00 = $197.73
The finance charge applied at the end of the third month is
$197.73 × .199/12 = $3.28
so Marilyn can make one final payment of
$197.73 +3.28 = $201.01
to pay off the balance.
In all, Marilyn has paid 2×$410.00 +201.01 =
$1021.01 . . . . . . . . corresponds to the first choice
_____
In real life, Marilyn's credit card may not accrue any finance charge until after the first statement on which the charge appears. Thus the total cost of the purchase may be only $1004.02. The attached spreadsheet shows the beginning balance and the finance charges for each month for the two different scenarios.
$991.38 × 0.199/12 = $16.44
The balance subject to the next month's finance charge will be
$991.38 +16.44 -410.00 = $597.82
The finance charge at the end of the second month will be
$597.82 × 0.199/12 = $9.91
The balance remaining after the second payment will be
$597.82 +9.91 -410.00 = $197.73
The finance charge applied at the end of the third month is
$197.73 × .199/12 = $3.28
so Marilyn can make one final payment of
$197.73 +3.28 = $201.01
to pay off the balance.
In all, Marilyn has paid 2×$410.00 +201.01 =
$1021.01 . . . . . . . . corresponds to the first choice
_____
In real life, Marilyn's credit card may not accrue any finance charge until after the first statement on which the charge appears. Thus the total cost of the purchase may be only $1004.02. The attached spreadsheet shows the beginning balance and the finance charges for each month for the two different scenarios.