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(#M40012034) TCS question Keep an EYE

‎0,1,2,3 use these numbers for 4digit numbers and find the sum?

Asked In TCS Sw@ti.. (12 years ago)

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akshay sharma  (12 years ago)

total no of no's possble are: 18, as "0" cannot occur at 1st place.
(n-1)!*sum of digits * 1111
3!*6*1111=39996
but sum of no's formed when 0 is at 1st place should be subtracted,
39996 - (2!*6*111)
39996 - 1332
38664.
if any doubt, form 4 digit no's and add them, no's are:
1023,1032,1203,1230,1302,1320, 2013,2031,2103,2130,2301,2310, 3012,3021,3102,3120,3201,3210
add them,will result to: 38664

Is this solution Helpfull? Yes (41) | No (1)  

gaya3  (12 years ago)

the total combinations possible are 4!=24..if we write the combinations one after the other we get 24 rows and 4 columns..in each column we will hav 6 zeroes,6 ones,6 twos and 6 threes=> each column counts=(6*0+6*1+6*2+6*3)=36=>so overall for four column addition we get..36 36 36
+3 +3 +3
----------
39996
so,the ANSWER IS 3996

Is this solution Helpfull? Yes (4) | No (9)  

tanushri godha  (12 years ago)

Akshay:-why you mul. by 1111
please explain.......

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abcd  (12 years ago)

anybody tell how to solve this..

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k.selva kumari  (11 years ago)

a=1,c=3,z=26
1+3+26+0+1+2+3=36

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