PLEASE HELP ASAP! Consider the 720 permutations of the numbers 1, 2, 3, 4, 5, 6. In how many of these permutations do exactly 3 of the numbers end up in the positions where they started?

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.

step-by-step explanation:

this is not possible to solve with the info given.   you need at least an one angle somewhere.

i would say b) is your answer

sorry if im wrong

step-by-step explanation: