The correct option is D 54
Let the beggar be given x number of one rupee coins, y number of 50 paisa coins and z number of 25 paisa coins.
⇒x+0.5y+0.25z≥1
where x=0,1,2; y=0,1,2,3; z=0,1,2,3,4
⇒4x+2y+z≥4
For x=1,2, y and z can take any value.
Number of ways =2×4×5=40
For x=0,2y+z≥4
For y=0,z=4→1 way
For y=1,z∈{2,3,4}→3 ways
For y=2,z∈{0,1,2,3,4}→5 ways
For y=3,z∈{0,1,2,3,4}→5 ways
Total number of ways =40+1+3+5+5=54