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.
T=2n+3
considering n has no other values given
Answer:
C is false
Step-by-step explanation:
Example 3 is odd but 6 is equal.
3•6 is 18
18 is even
Answer:
What's the equation?
Step-by-step explanation:
We can write a proportion to resemble the problem;
AE/ED = AB/BC
AE = 9
ED = 6
AB = x
BC = 10
Substitute with the given values.
9/6 = x/10
9/6 * 10 = x/10 * 10
90/6 = x
15 = x
Therefore, the answer is 15.
Best of Luck!
