Question

Find the number of positive integers which can be formed by using any number of digits from $0,1,2,3,4,5$ but using each digit not more than once in each number. How many of these integers are greater than $3000$?

Answer 1

0, 1, 2, 3, 4, 5. Six digits : 5 are + ive and one zero.
We are at liberty to use any number of digits out of the six. Single digit numbers are clearly 5 which are + ive. ...(1)
Two digit numbers
${}^6{P}_2{-}^5{P}_1=\frac{6!}{4!}-5=30-5=25$ ...(2)
${}^5{P}_1$ corresponds to the two digit numbers having 0 in the first place.
Three digit numbers
${}^6{P}_3{-}^5{P}_2=\frac{6!}{3!}-\frac{5!}{3!}=\frac{720-120}{6}=100.$ ...(3)
Four digit numbers
${}^6{P}_4{-}^5{P}_3=\frac{6!}{2!}-\frac{5!}{2!}=\frac{720-120}{2}=300.$ ...(4)
Five digit numbers
${}^6{P}_5{-}^5{P}_4=6!-5!=720-120=600.$ ...(5)
Six digit numbers
${}^6{P}_6{-}^5{P}_5=6!-5!=720-120=600.$ ...(6)
Total = 600 + 600 + 300 + 100 + 25 + 5 = 1630.
2nd part. Numbers greater than 3000.
All the five digit numbers 600 and six digit numbers 600 will be greater than 3000 by (5) and (6).
All the four digit numbers with either 3 or 4 or 5 in the first place will be greater than 3000.
Numbers of 4 digit having 3 in the first place will be
${}^{5}P3=\frac{5!}{2!}=\frac{120}{2}=60$
Similarly those having 4 and 5 in the first place. Therefore 4 digit numbers greater than 3000 is
60 + 60 + 60 = 180.
Hence total numbers greater than 3000 will be $600+600+180=1380.$