Answer:
Step-by-step explanation:
The set {1,2,3,4,5,6} has a total of 6! permutations
a. Of those 6! permutations, 5!=120 begin with 1. So first 120 numbers would contain 1 as the unit digit.
b. The next 120, including the 124th, would begin with '2'
c. Then of the 5! numbers beginning with 2, there are 4!=24 including the 124th number, which have the second digit =1
d. Of these 4! permutations beginning with 21, there are 3!=6 including the 124th permutation which have third digit 3
e. Among these 3! permutations beginning with 213, there are 2 numbers with the fourth digit =4 (121th & 122th), 2 with fourth digit 5 (numbers 123 & 124) and 2 with fourth digit 6 (numbers 125 and 126).
Lastly, of the 2! permutations beginning with 2135, there is one with 5th digit 4 (number 123) and one with 5 digit 6 (number 124).
∴ The 124th number is 213564
Similarly reversing the above procedure we can determine the position of 321546 to be 267th on the list.
Answer:
6
Step-by-step explanation:
Answer:
A = 219
C = 362
Step-by-step explanation:
Given,
581 = C + A
and
$1090.50 = 1.5C + 2.5A
solve 1st for A
A =581 - C
substute into 2nd
1090.50 = 1.5C + 2.5(581 - C)
solve for C
1090.50 = 1.5C + 1452.50 - 2.5C
1090.50 = -C + 1452.50
1090.50 - 1452.50 = -C
-362 = -C
C = 362
then
581 = C + A
581 = 362 + A
219 = A
A = 219
Answer: $65.50 - 15% = 55.675
Step-by-step explanation:
Conversion a decimal number to a fraction: 65.50 = 655
10
= 131
2
Conversion a decimal number to a fraction: 9.825 = 9825
1000
= 393
40
Subtract: 65.5 - 9.825 = 131
2
- 393
40
= 131 · 20
2 · 20
- 393
40
= 2620
40
- 393
40
= 2620 - 393
40
= 2227
40
The common denominator you can calculate as the least common multiple of the both denominators - LCM(2, 40) = 40
Answer:
- solutions: (2, 10), (4, 5)
- non-solutions: all other points
Step-by-step explanation:
It can be useful to graph the inequalities related to Tyler's limits of money and weight.
1x +3y ≤ 36 . . . . . . the limit on the cost of the items Tyler can afford
1x +1.5y ≤ 20 . . . . . the limit on the weight of the items Tyler can carry
Then plotting the given points shows that only (2, 10) and (4, 5) are in the doubly-shaded area that is the solution space. The other points are not a solution.
