Question

How many numbers can be formed with $1,3,5$ and $7$ when no digit is repeated in any number?

Answer 1

$1,3,5,7$

No of one digit number =4

No. of two digit numbers

$=4P_2$

$=\frac{4!}{(4-2)!}$

$=4\times 3$

$=12$

No. of three digit numbers

$=4P_3$

$=\frac{4!}{(4-3)!}$

$=4\times 3\times 2$

$=24$

No. of 4-digit numbers

$=4P_4$

$=\frac{4!}{(4-4)!}$

$=4\times 3\times 2$

$=24$

Total number of ways $=12+24+24+4$

$=64$