Latest TCS Aptitude Question SOLUTION: How many 4 digit no's can be formed using the digits 0-9 with repetition of digits allowed such that 3 appears exactly twice?

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How many 4 digit no's can be formed using the digits 0-9 with repetition of digits allowed such that 3 appears exactly twice?

Asked In TCS Kunal Verma (10 years ago)

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rohan chawla  (10 years ago)

answer is 459
here we have a 4 digit number so msb or left most digit cannot be 0
now we take two cases

case 1
when 3 is in the msb
so in the remaining 3 places we require one 3 and two numbers from 0 to 9 except 3 i.e. 9 numbers
so 3 can be put in 3 places in 3 ways and remaining two places can be filled in 9*9 = 81 ways
so total number of ways are 81*3 = 243 ways

case 2
here there is no 3 in msb. both 3 are in remaining three places
it means msb can be filled by numbers from 0 to 9 except 0 and 3 that is in 8 ways
and the remaining three places can be filled by two 3 and any of the numbers from 0 to 9 except 3 that is 9 ways
two 3 can be filled in three places in 3 ways
so number of arrangements are 8*9*3 = 216

so total number of ways is 216+243 = 459

Is this solution Helpfull? Yes (28) | No (2)  

Kunal Verma  (10 years ago)

Ans-459
possible ways can be 33xx xx33 3x3x x3x3 x33x 3xx3 where x can be 0-9 except 3 so 9*9=81 for 1 way so similarly 4 six ways 6*81= 486 - the cases where 0 is at first position

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vikrant  (10 years ago)

ans is 459
here two case will arise
1)3 is present at thousnd's place
2)3 is nt there
total no of arranging twice 3 at 4 places=4C2=6
from thease 6,3 will contain 3 at thousand's place
so in this case total1=(1*9*9*1)*3=243
and for rest 3
total2=(8*9*1*1)*3=216
so total=216+243=259

Is this solution Helpfull? Yes (4) | No   

nikhil pande  (10 years ago)

9*1*1*10+1*1*10*10+9*1*10*1+1*10*10*1=380

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