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Numerical Ability
Permutation and Combination
A person starts writing all 4 digit numbers .How many times had he written the digit 3?
Read Solution (Total 15)
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- when 3 is at unit place= ---3 => 9*10*10=900
when 3 is at tenth place= --3- => 9*10*10=900
when 3 is at hundred place= -3-- => 9*10*10=900
when 3 is at thousand place= 3--- => 10*10*10=1000
total no. of 3's = 900+900+900+1000 = 3700
- 10 years agoHelpfull: Yes(52) No(18)
- 4 digit no ....
let start with unit digit is 3 then [][][][3]
4th 3rd 2nd
4th position is filled in 9 ways 1,2,3,.....9.except 0;because when we put 0 at 4th position then number would be of 3 digit.
now 3rd position is filled in 10 ways by 0,1,2.....9.
now 2nd position is also filled in 10 ways.
so total possible ways=9*10*10*1;
similarly,
let 2nd digit is 3 i.e.[][][3][]
so 4th position digit is filled in 9 ways
3rd...............................9 ways
2nd position .......................in 1 way i.e 3 we have already filled.
1st position is filled in 10 ways
so total ways=9*10*1*10
similarly we put 3 at 3rd position then we get 9*1*10*10
now,
when we put 3 at 4th position [3][][][]
then 1st position is filled in 1 way i.e 1 then 2nd position in 10 ways i.e
0,1,2.......9 now 3rd position is also filled in 10 ways and 4th position is also filled in 10 ways ......so total possible ways=1*10*10*10.
now total ways=9*10*10*1+9*10*1*10+9*1*10*10+1*10*10*10=3700 - 10 years agoHelpfull: Yes(14) No(2)
- when 3 is at unit,tens, hundredth & thousands positions
---3 => 9*10*10 = 9000
--3- => 9*10*10 = 9000
-3-- => 9*10*10 = 9000
3--- => 10*10*10 = 1000
Total no. of 3's = 900+900+900+1000 = 3700 - 10 years agoHelpfull: Yes(10) No(13)
- Hey Rakesh can u plz explain 9*10*10?
- 10 years agoHelpfull: Yes(4) No(4)
- ans is 3168
total no of 4 digit no.s = 9*10*10*10=9000
no. of 4 digit no. which not having 2 = 8*9*9*9 = 5832
no. which havig 2 is = 9000-5832=3168 - 10 years agoHelpfull: Yes(3) No(3)
- In 1 to 100 it appears=20 times
In 1 to 1000 it appears=10*20+100=300
In 1 to 10000 it appears=10*300+1000=4000
So ans is 4000 - 10 years agoHelpfull: Yes(3) No(0)
- 3700 is wrong, since the case in which two digits are 3 such as 3391 is not taken into account. correct ans is- 3168.
total no of 4 digit no.s = 9*10*10*10=9000
no. of 4 digit no. which not having 3 = 8*9*9*9 = 5832
no. which is having 3 is = 9000-5832=3168 - 10 years agoHelpfull: Yes(2) No(1)
- [][][][3] :first box can be filled in 9 ways(except 0 otherwise it will be 3 digit).second and third box can be filled in 10 ways respectively.so total 9x10x10ways.
[][][3][]: Same as above 9x10x1x10 ways.
[][3][][]: Same 9x1x10x10 ways
[3][][][]: 2nd,3rd,4th box can be filled in 10x10x10 ways.
So, total ways of forming four digit number will be :sum off all the 4 above cases=3700.
- 10 years agoHelpfull: Yes(2) No(0)
- assume that the numbers are 0,1,2,3,4,5,6,7,8,9 total 10.
now first place the digit 3 in unit's place i.e,the first place can be filled in 1 way similarly 2nd place can be filled with rest of the 9 numbers,3rd place in 8 ways,4th place(thousand's place) in 7 ways. so total number of ways for 3 placed in unit's place is
1*9*8*7=504 there are total 4 places so place your 3 in hundreds,ten's&thousand's places also and finally we get the answer as
4(1*9*8*7)=2016. - 10 years agoHelpfull: Yes(1) No(6)
- Hey Rakesh can u plz explain 9*10*10?
- 10 years agoHelpfull: Yes(1) No(1)
- the ans is 3997......
- 10 years agoHelpfull: Yes(1) No(0)
- we have to make 4 digit number that is to fill 4 vacant places _ _ _ _.CASE 1=when only 1 three appears on the vacant place .CASE 2=when only 2 three appears. CASE 3=when only 3 three appears .CASE 4=when all 4 digits are three.take permutation of all cases we get 3321 as the answer.
- 10 years agoHelpfull: Yes(0) No(2)
- when 3 is in unit place it can be arranged in only one way , remaing places (ten's place, hundered place can be arranged in 10 ways(0,1,2,.....9) where as thousands place must be arranged in 9 ways only (bcoz it is four digit number so, it can be arranged from 1,2,..9)
- 10 years agoHelpfull: Yes(0) No(0)
- hiii RAKESH,Can u plz explain in detail....
- 10 years agoHelpfull: Yes(0) No(0)
- We use 20 times 3's in 1-100 nos. And in 300-400 119 times so totally in 1000-1999 we have used 20*9+119=299 times so similarly for 2000-2999 so on except 3000-3999 so till now 299*9=2691.now in 3000-3999 we use 1000 more 3's.
Total 2691+1000=3691 - 10 years agoHelpfull: Yes(0) No(0)
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