Question:
How many numbers can be formed from the digits 1, 3, 5, 9 if repetition of digits is not allowed?
Solution:
To find: number of numbers that can be formed from the digits 1, 3, 5, 9 if repetition of digits is not allowed
Forming a 4 digit number: $4 !$
Forming a 3 digit number: ${ }^{4} \mathrm{C}_{3} \times 3 !$
Forming a 2 digit number: ${ }^{4} \mathrm{C}_{2} \times 2 !$
Forming a 1 digit number: 4
So total number of ways $=4 !+\left({ }^{4} \mathrm{C}_{3} \times 3 !\right)+\left({ }^{4} \mathrm{C}_{2} \times 2 !\right)+4$
$=24+24+12+4$
$=64$