TCS
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Numerical Ability
Permutation and Combination
find the sum of all four digit numbers that can be formed using the digits 0,1,2,3. no digits can be repeated.
Read Solution (Total 16)
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- From these four digits '0' cannot be at first place to form four digit number so we have 1, 2, or 3 can be at thousand's place keeping one digit at thousand place can have 6 different four digit number so that means 1000*6 + 2000* 6 + 3000* 6 = 36000.
Now come to hundred's place we can have all four digit at this place , '0' will not contribute any thing to sum so in all four digits number 1,2,3 will be at 4 times on same place so that means 100*4 + 200* 4 + 300 * 4= 2400.
Now come to tenth's place ...same as hundred so 10*4 + 20*4 + 30*4= 240.
same for unit place 1*4 + 2*4 + 3*4= 24 in last if you add whole the total sum of all possible four digit number formed is 38664 - 10 years agoHelpfull: Yes(26) No(3)
- (n-1)!*1111*(sum)
(4-1)!*1111*6
39996 - 10 years agoHelpfull: Yes(25) No(4)
- three cases-
when 1 in thousands place=
1 0 2 3
1 0 3 2
1 2 0 3
1 2 3 0
1 3 0 2
1 3 2 0
similarly for 2 and 3 in thousands place- 6-6 cases each
total num=6+6+6=18
if we see, in units, tens and hundreds place-
4 times 1=4
4 times 2=8
4 times 3=12
6 times 0=0
so sum at units,tens and hundreds place=4+8+12=24, 24, 24
thousandth place-
6 times 1=6
6 times 2=12
6 times 3=18
sum at thousnds=6+12+18=36
now sum = 36*1000 + 24*100 + 24*10 + 24*1 = 38664
- 10 years agoHelpfull: Yes(20) No(3)
- Sum of all the r-digits formed by taking the given n-digits = (sum of all n-digits ).n-1Pr-1*(1111.....r times)
So here n=4 and r=4 - 10 years agoHelpfull: Yes(3) No(5)
- total non repeated numbers:3*3*2*1=18
sum of all numbers will be:(1000)*6*18+(100)*6*18+(10)*6*18+(1)*6*18
=108*(1111)
=119988 - 10 years agoHelpfull: Yes(2) No(1)
- since 0 is present the formula becomes
(n-1)!*(1111)(sum of digits)-(n-2)!*1111(Sum of digits)
1111*7*(6-2)=31108 - 10 years agoHelpfull: Yes(2) No(0)
- Any way and = 0+1+2+3=6
- 10 years agoHelpfull: Yes(1) No(2)
- as there is zero in the given nos the formula is ((n-1)!*(1111)*(0+1+2+3))-((n-2)!*(1111)*(0+1+2+3)=26664
- 10 years agoHelpfull: Yes(1) No(0)
- 4! i.e equals to 4x3x2x1=24
- 10 years agoHelpfull: Yes(1) No(2)
- For thousadth position=1000(1+2+3+0)*3*4*4*4
=1000(6)*192
For hundredth position sum=100(6)192
For Tenth position sum=10(6)*192
Unit position sum=1(6)8192
overall sum=(6)192(1000+100+10+1) =1279872 - 10 years agoHelpfull: Yes(0) No(7)
- lowest 4 didgit no :1000
highest 4 digit no :3333
so sum on all numbers from 1000-3333 = (3333*3334)/2-(999*1000)/2
here we are finding sum of numbers from 1-3333 and then removing from 1-999
Formula = (n*n+1)/2 for sum of n numbers - 10 years agoHelpfull: Yes(0) No(1)
- no 0 is allowed in thousend position
so ans=>38664 - 10 years agoHelpfull: Yes(0) No(0)
- answer is 56664
- 10 years agoHelpfull: Yes(0) No(0)
- no digit repeated so,at every place 4 numbers in 4 ways...
so 4*4*4*4=256 ...i think - 10 years agoHelpfull: Yes(0) No(3)
- 38664... as there would be 18 different arrangements with taking number 1 , 2,3 in thousands place.... adding these all no would result in no 38664
- 10 years agoHelpfull: Yes(0) No(0)
- plzzz tell how many pages i have to solve so that iam able to solve tcs one....
in set of 30 quesytion - 10 years agoHelpfull: Yes(0) No(0)
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