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Numerical Ability
Arithmetic
How many positive integers less than 450 can be formed using 1,2,5 and 6 for it's digits, with each digit being used only once?
Read Solution (Total 12)
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- Ans: 28
Num of possible 1 digit numbers = 4
Num of possible 2 digit numbers = 4*3 = 12
For 3 digit numbers, to meet the condition
the number should be less than 450, 5 and 6 cannot be in the hundredth
position.
Num of possible 3 digit numbers = 2*3*2 = 12
Total +ve integers meeting the condition = 4+12+12 = 28 - 11 years agoHelpfull: Yes(62) No(6)
- yes it will be 28 numbers.... i made a mistake...
- 11 years agoHelpfull: Yes(8) No(3)
- three digit no=2*3*2=12
two digit no=4*3=12
one digit no=4
so, total nos= 12+12+4=28 ANS - 11 years agoHelpfull: Yes(4) No(0)
- 12 such integers are possible
- 11 years agoHelpfull: Yes(2) No(7)
- guys will u plz xplain me why is it 2*3*2 for 3digit no.?
- 10 years agoHelpfull: Yes(2) No(1)
- @ Prakash: Yes,Many made the same mistake in a test,including me!
- 11 years agoHelpfull: Yes(1) No(1)
- @karthika- yup!! we get the perception that a 3 digit no less than 450 is required..
- 11 years agoHelpfull: Yes(1) No(1)
- no of integers should be 28.
1 digit no's=4
2 digit no's be=4*3=12
3 digit no's be=2*3*2=12 - 11 years agoHelpfull: Yes(1) No(0)
- Yeah! yu r ryt :D
- 11 years agoHelpfull: Yes(0) No(1)
- 16+4+12=32
- 11 years agoHelpfull: Yes(0) No(9)
- 1 digit number = 4
2 digit number = 4*3 =12
3 digit number< 450 =2*3*2=12
adding three =28 - 11 years agoHelpfull: Yes(0) No(0)
- can anyone explain . Num of possible 2 digit numbers = 4*3 = 12
where does the 3 come from ? and what is it? - 10 years agoHelpfull: Yes(0) No(0)
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