Question

There are 720 permutations of the digits 1,2,3,4,5,6. suppose these permutations are arranged from smallest to largest numerical of values, beginning from 1 2 3 4 5 6 and ending with 6 5 4 3 2 1. (a)what number falls on ${124}^{th}$ position? (b) what is the position of 321546?

Answer 1

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 $124$th, would begin with ${}^{\prime }{2}^{\prime }$

c. Then of the $5!$ numbers beginning with $2$, there are $4!=24$ including the $124$th number, which have the second digit $=1$

d. Of these $4!$ permutations beginning with $21$, there are $3!=6$ including the $124$th permutation which have third digit $3$

e. Among these $3!$ permutations beginning with $213$, there are $2$ numbers with the fourth digit $=4$ ($121$th & $122$th), $2$ 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 $5$th digit $4$ (number $123$) and one with $5$ digit $6$ (number $124$).

$\therefore$ The $124$th number is $213564$

Similarly reversing the above procedure we can determine the position of $321546$ to be $267$th on the list.