How many natural numbers not more than 4300 can be formed with the digits 0,1,2,3,4 (if repetitions are not allowed)?

My Approach:

Total no of available digits: 5

No of 1-digit formed: 5

No of 2-digit formed: 4(excluding '0' on ten's place) * 4 (including zero - one among given digits already used)

No of 3-digit formed: 4(excluding '0' on 100th place) * 4 (including zero - one among given digits already used) * 3

No of 4-digit formed less than 4000: 3(excluding '0' & '4' on 1000th place) * 4 (including '0' & '4') * 3 * 2

No of 4-digit formed less than 4300 and greater than 4000: 3(excluding '3' & '4' on 100th place) * 3 (including '3') * 2

5 + (4*4) + (4*4*3) + (3*4*3*2) + (3*3*2)

So total of 175 natural numbers can be formed.

But answer is 158. Please explain how approach this kind of problems

• $5+4*4+4*4*3+3*4*3*2+3*3*2=5+4^3+5*3*3*2=5+64+90=159$, I guess they don't consider $0$ to be natural. – blues66 Mar 18 '14 at 3:05

Your calculations are correct, but $5 + (4*4) + (4*4*3) + (3*4*3*2) + (3*3*2)=159$,
not $175$, as blues66 suggested they probably didn't include $0$ as a solution