Latest TCS Aptitude Question SOLUTION: find the sum of all four digits numbers that can be formed using the digits 0,1,2,3. no digits can be repeated.

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(#M40039004) TCS Numerical Ability question number theory Keep an EYE

find the sum of all four digits numbers that can be formed using the digits 0,1,2,3. no digits can be repeated.

Asked In TCS Thamizhselvam (9 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

pritika singh  (9 years ago)

sum of all no. formed=(n-1)!*(sum of n digits)*(111..n times)
=3!*(6)*(1111)
=39996

Is this solution Helpfull? Yes (20) | No (1)  

SATYA MADHURI  (9 years ago)

Sum of all no. formed that are distinct = (n-1)! * (sum of n digits) * (111.. n times)
= 3! * (0+1+2+3) * (1111)
= 39,996
The above includes when 0 is in thousands place.

Hence we need to subtract that value, which means we need to find the sum of all the numbers formed using 1,2,3
Using the above formula we get
=2! * (3+2+1) * (111)
=1332

Hence 39,996-1,332 = 38,664

38,664 is the answer

Is this solution Helpfull? Yes (10) | No   

MUNISH YADAV  (9 years ago)

3012
3021
3120
3102
3201
3210
Sum is 18666
2310
2301
2031
2013
2103
2130
Sum is 12888
1320
1302
1023
1032
1203
1230
Sum is 7110
Total sum = 18666+12888+7110=38664

Is this solution Helpfull? Yes (8) | No (1)  

Loganathan J.  (9 years ago)

1+2+3+0=6
(4-1)!
1111
6*3!*1111=39996

Is this solution Helpfull? Yes (4) | No (2)  

abhyuday pal  (9 years ago)

3!*1000(3+2+1) + 2*2*1*100(3+2+1) + 2*2*1*10(3+2+1) + 2*2*1*(3+2+1) { + 3!*100(0) + 3!*10(0) + 3!(0)} = 38664
PS: the term inside the braces =0 . Added just for better understanding & explanation

Is this solution Helpfull? Yes (4) | No   

Loganathan J.  (9 years ago)

3*3*2*1=18

Is this solution Helpfull? Yes (1) | No (1)  

puja agarwal  (9 years ago)

(0+1+2+3)(1111)(6)-(0+1+2+3)(111)(2)
=38664

Is this solution Helpfull? Yes (1) | No (1)  

gautam gaurav  (9 years ago)

total sum=4(3)(1111)(6)=79992

Is this solution Helpfull? Yes | No (2)  

Bharti jain  (9 years ago)

Munish yadav..
is there any shortcut method.

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