Question

All the $7-$digit numbers containing each of the digits $1,2,3,4,5,6,7$ exactly once, and not divisible by $5$, are arranged in increasing order. Then which among the following statements is INCORRECT $?$

• A. ${1993}^{\text{rd}}$ number in the list is $4312567$
• B. ${1996}^{\text{th}}$ number in the list is $4312756$
• C. ${2000}^{\text{th}}$ number in the list is $4315672$
• D. ${1999}^{\text{th}}$ number in the list is $4315276$

Answer 1

The correct option is D ${1999}^{\text{th}}$ number in the list is $4315276$

The $7-$digit number with $1$ in the left most place and containing each of the digits $1,2,3,4,5,6,7$ exactly once is $6!$

But $5!$ numbers out of those $6!$ have $5$ at the end that are divisible by $5.$
Hence number of numbers starting with $1\_\_\_\_\_\_=6!-5!=720-120=600$
number of numbers starting with $2\_\_\_\_\_\_=600$
number of numbers starting with $2\_\_\_\_\_\_=600$
total $=1800$

number of numbers starting with $41\_\_\_\_\_=120-24=96$
total $1800+96=1896$

number of numbers starting with $42\_\_\_\_\_=120-24=96$
total $1896+96=1992$

number of numbers starting with $43\_\_\_\_\_=120-24=96$
total $1992+96=2088$

So, ${1993}^{\text{th}}$ number is $4312567$
Next $7$ numbers after $4312567$ are :
$4312576,4312657,4312756,4315267,4315276,4315627,4315672.$