HELP? Consider the 720 permutations of the numbers 1, 2, 3, 4, 5, 6. In how many of these permutations does 1 appear next to 2 or 3 appear next to 4 or 5 appear next to 6?

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.

no pictures of the graph

your answer is a

step-by-step explanation:

4*4*4*4=256