Question

There are 720 permutations of the digits 1, 2, 3, 4, 5, 6. Suppose these permutations are arranged from smallest to largest numerical values, beginning from 123456 to 654321.
(a) What number falls on ${{124}^{th}}$ position?
(b) What is the position of 321546?

Answer 1

We solve this problem using the permutations. This problem is more similar to arranging the words in a dictionary. We assume that each digit is an alphabet in that order. Then arranging the numbers in ascending order is nothing but arranging the words in a dictionary.
We use one standard result of permutation that is the number of ways of arranging $'n'$ objects in $'n'$ places is given as $n!$
We take 6 boxes and place numbers in 6 boxes in an order of ascending order to find the number in ${{124}^{th}}$ position and we use the same technique until we get the number 321546 to find its position.

Complete step by step answer:
(a) Let us find the number in ${{124}^{th}}$ position.
Let us assume that there are 6 boxes to be filled with given 6 digits.

Now, let us assume that the first box is filled with 1 because we need to go in ascending order.

Now, we have 5 digits to be filled in remaining 5 boxes
We know that the number of ways of arranging $'n'$ objects in $'n'$ places is given as $n!$
By using the above formula the number of ways of arranging 5 digits in 5 boxes is
$\Rightarrow 5!=120$
So, we can say that the largest number having the 1 as first digit occupies ${{120}^{th}}$ position
Therefore the ${{120}^{th}}$ number is 165432
Now, let us place 2 in first box and write the numbers in ascending order then we get

Now the number of ways of arranging the remaining numbers is
$\Rightarrow 5!=120$
Now we get the total number of numbers are
$\Rightarrow 120+120=240$
But we need only ${{124}^{th}}$ number
So, we can say that digit 2 is fixed in first position
Now let us place next three digits in ascending order as fixed that is

Now, we get the number of ways of arranging the two digits in two places is
$\Rightarrow 2!=2$
So, we can have the total number of numbers until now as
$\Rightarrow 120+2=122$
So, we can say that the largest number after fixing 4 digits as shown above will have ${{122}^{nd}}$ position that is 213465
Now let us fix first three as it is and change the fourth digit to next number that is

Now, we get the number of ways of arranging the two digits in two places is
$\Rightarrow 2!=2$
So, we can have the total number of numbers until now as
$\Rightarrow 122+2=124$
So, we can say that the largest number after fixing 4 digits as shown above will have ${{124}^{th}}$ position that is 213564
Therefore the number in ${{124}^{th}}$ position is $213564$
(b) Position of 321546
Let us assume that there are 6 boxes to be filled with given 6 digits.

Now, let us assume that the first box is filled with 1 because we need to go in ascending order.

Now, we have 5 digits to be filled in remaining 5 boxes
We know that the number of ways of arranging $'n'$ objects in $'n'$ places is given as $n!$
By using the above formula the number of ways of arranging 5 digits in 5 boxes is
$\Rightarrow 5!=120$
Now, let us place 2 in first box and write the numbers in ascending order then we get

Now the number of ways of arranging the remaining numbers is
$\Rightarrow 5!=120$
Now we get the total number of numbers are
$\Rightarrow 120+120=240$
Now let us place 3 in first box then we get

We are asked to find position of 321546 so we can leave the first position to 3 and let us place least number from the remaining digits that is 1 in second box as

Now the number of ways of arranging the remaining numbers is
$\Rightarrow 4!=24$
Now we get the total number of numbers are
$\Rightarrow 240+24=264$
Now let us place 2 in second box as

We are asked to find position of 321546 so we can leave the first position to 3 and second position to 2 and let us place least number from the remaining digits that is 1 in third box as

We are asked to find the position of 321546 so we can leave the number positions as it is shown in the above figure.
Now, let us place least number from remaining digits that is 4 in fourth place

Now the number of ways of arranging the remaining numbers is
$\Rightarrow 2!=2$
Now we get the total number of numbers are
$\Rightarrow 264+2=266$
Now let us place 5 in fourth place that is

We are asked to find the position of 321546 so we can leave the number positions as it is shown in the above figure.
Now, let us place least number from remaining digits that is 4 in fifth place

Now the number of ways of arranging the remaining numbers is
$\Rightarrow 1!=1$
Now we get the total number of numbers are
$\Rightarrow 266+1=267$
We are asked to find the position of 321546 so we can leave the number positions as it is shown in the above figure.
Now, let us place least number from remaining digits that is 6 in sixth place

This arrangement can be done in one way.
Now we get the total number of numbers are
$\Rightarrow 267+1=268$

Therefore, the position of number $321546$ is $268$.

Note: Students may make mistakes in considering the permutations.
We are given that permutations are arranged from smallest to largest numerical values, beginning from 123456 to 654321
Students may think that this an Arithmetic progression with first term as 123456 and common difference as 1
This gives the wrong answer because the digits 7, 8, 9, 0 are not there in the permutations then this assumption of A.P will be completely wrong.
Also in the second part we have the total number of numbers after fixing 4 digits as
$\Rightarrow 266+1=267$
Then after fixing the last two digits we get the number of numbers up to 321546 as
$\Rightarrow 267+1=268$
But students may make mistakes without adding 1 to 267 and give the position of 321546 as 267 which will be wrong because we need to add the position of 321546 to get the final answer.