Question

I. The number of all ten digited numbers that can be formed with all the distinct digits and which are divisible by $4$ is $15\times 8!$.
II. The number of positive integers that can be formed by using the digits $0,1,2,3,4,5$ without any repetition is $1630$.

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Both I and II are false

Answer 1

Answer: (B)

I. We have,
$1,2,3,4,5,6,7,8,9$
The number formed can be divisible by $4$ when last two digits has to be divisible by $4$ .
There are $16$ last two digits numbers that are divisible by $4$ without using zero.
So, total number of such $10$ digits numbers $=7\times 7!\times 16$
Again, there are $6$ last two digits numbers that are divisible by $4$ with using zero.
So, total number of such $10$ digit numbers $=8!\times 6$
$\therefore$ Required numbers
$=7\times 7!\times 16+8!\times 6$
$=7\times 8!\times 2+8!\times 6$
$=8!(14+6)=20\times 8!$
So, statements I is wrong.
II. Six digit numbers $=5\times 5\times 4\times 3\times 2\times 1=600$
Five digit numbers $=5\times 5\times 4\times 3\times 2=600$
Four digit numbers $=5\times 5\times 4\times 3=300$
Three digit numbers $=5\times 5\times 4=100$
Two digit numbers $=5\times 5=25$
One digit number $=5=5$
$\therefore$ Required number $=600+600+300+100+25+5=1630$