UPSC NDA II 2015 Mathematics Solved Paper

No. of digits to be filled at one’s place =3
No. of digits to be filled at 10’s place =5
No. of digits to be filled at 100’s place=4
∴Total no. of digits formed=3×5×4=60
If zero is at 100’s place;
Then; no. of digits to be filled at one’s place=2
& no. of digits to be filled at 10’s place=4
∴No. of digits formed with zero at 100’s place=1×2×4=8
∴Required no. of digits formed=60−8=52.