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

sum of all no. formed=(n-1)!*(sum of n digits)*(111..n times)
=3!*(6)*(1111)
=39996
10 years agoHelpfull: Yes(20) No(1)

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
10 years agoHelpfull: Yes(10) No(0)
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
10 years agoHelpfull: Yes(8) No(1)
1+2+3+0=6
(4-1)!
1111
6*3!*1111=39996
10 years agoHelpfull: Yes(4) No(2)
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
10 years agoHelpfull: Yes(4) No(0)
3*3*2*1=18
10 years agoHelpfull: Yes(1) No(1)
(0+1+2+3)(1111)(6)-(0+1+2+3)(111)(2)
=38664
10 years agoHelpfull: Yes(1) No(1)
total sum=4(3)(1111)(6)=79992
10 years agoHelpfull: Yes(0) No(2)
Munish yadav..
is there any shortcut method.
10 years agoHelpfull: Yes(0) No(0)

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