Question

How many numbers which are
(i) Even
(ii) Less than 40,000 can be formed by taking all the digits 1, 2, 3, 4, 5?
(iii) The number of odd numbers between 1000 and 10,000 can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 is

• A. $1280$
• B. $1836$
• C. $2572$
• D. $1680$

Answer 1

Answer: (D)

$1680$

$1,2,3,4,5:5$ digits

(i) Even numbers which will have $2$ in the last place = $4!=24.$

Similarly those which will have $4$ in the last place =$4!=24$
Total is $24+24=48$.

(ii) Numbers less than $40,000$ will have either $1or2or3$ in the first place and

zhence as above total of such numbers will be $24+24+24=72$.

(iii) Ans. (d). It will be $4$ digit number and as it is to be odd the unit place can be filled in $5$ ways by any of the $5$ odd numbers. Out of remaining $8$ we have to arrange $3$ in ${}^8{P}_3$ ways.

$(8\times 7\times 6)\times 5=1680$ by fundamental theorem.