How many numbers can be formed by using digits 0 to 9 as the number should be divisible by 5

8p3 for 1st 3 digits =336
2p1 for last digit =2
result =336*2 =672
10 years agoHelpfull: Yes(11) No(12)
_ _ _ _:4th digit 0 or 5 so , 2P1=2
3rd digit has 9 ways i.e) 9P1=9
2nd digit has 8 ways i.e) 8P1=8
1st digit has 7 ways i.e) 7P1=7
result=9*8*7*2=1008
9 years agoHelpfull: Yes(4) No(3)
Total numbers can be formed are==> 9*9*8*7
9*9*8*7-((9*8*7)+(8*8*7))
10 years agoHelpfull: Yes(3) No(1)
frds.dont get confused.praveen kumar is absoluetely correct.nice ans praveen..
9 years agoHelpfull: Yes(3) No(0)
9p3 ie...9*8*7=504
10 years agoHelpfull: Yes(1) No(5)
630 numbers can be formed
9 years agoHelpfull: Yes(0) No(1)
4-digits->!@#$
$ can have oly 2 ways (0 or 5) = 2
! can have 8 ways since we have taken off 0 and 5 hence we have 8
@ can have 7 ways
# can have 6 ways

Hence we have 8*7*6*2 ways = 672 ways
9 years agoHelpfull: Yes(0) No(3)
ANS : 952
4 digit number from 1000 to 9999
4th digit must be 0/5 So 2 possibilities
If 4th digit is 0
1st as 9, 2nd has 8 and 3rd digit has 7 possibilities
If 4th digit is 5
1st as 8(0 would turn it into 3 digit number), 2nd has 8 and 3rd digit has 7 possibilities

So possibilities = 9*8*7+8*8*7=56*17=840+112=952
9 years agoHelpfull: Yes(0) No(0)
unit digit can be 5 or 0 : 2 ways
rest 3 ways can be filled in 9*8*7
ans 9*8*7*2=1008
9 years agoHelpfull: Yes(0) No(0)

How many numbers can be formed by using digits 0 to 9, as the number should be divisible by 5 and it should be 4 digit one and no repitation is allowed on digits?