Latest TCS Aptitude Question SOLUTION: Three variables x,y,z have sum of 30.All three of them are non-negetive integers.If any two variables don't have same value and extacly one variable has a value less han or equal t

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Three variables x,y,z have sum of 30.All three of them are non-negetive integers.If any two variables don't have same value and extacly one variable has a value less han or equal to three,then find the number of possible solution for variables.

Asked In TCS Gaurav Pratap Singh (8 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

prashanthi  (8 years ago)

let digits be X,Y,Z
X Y Z
0 (1 ,29),(2,28),(3,27) ............(29,1) no of possibilities=29
1 (0,29),(2,27),(3,26).............(27,2),(29,0) no of possibilities=28
2 (0,28),(1,27),(3,25).............(25,3),(27,1),(28,0) no of possibilities=27
3 (0,27),(1,26),(2,25),(4,23)...........(23,4),(25,2),(26,1),(27,0) no of possibilities=26
total no of possible solutions are 29+28+27+26=110 possibilities

Is this solution Helpfull? Yes (17) | No (11)  

Abhra Pratim Chowdhury  (8 years ago)

A= 3 there is 20possiblities of selecting different numbers of b and c
A= 2 , -----------------------20possiblities
A=1 ____________ 22posiblities
A=0 again 22posiblities so answer is 20+ 20+ 22+ 22= 84

Is this solution Helpfull? Yes (5) | No (10)  

shahbaz khan  (8 years ago)

solution 2 must be modified.
The number of ways of selecting exactly one variable having value less than or equal to 3 is 3.
So the number of solution is=3*84=252

Is this solution Helpfull? Yes (1) | No (3)  

B.Nikitha  (8 years ago)

let x,y,z are the digits
z has 4 possibilities i.e., z=0,1,2,3
so wen z=0 there are 28 possibilities
when z=1 there are 26 possibilities
when z=2 there are 22 possibilities
when z=3 there are 20
so total 96

Is this solution Helpfull? Yes | No (4)  

richa sharma  (8 years ago)

which one is correct plzzztell me?//

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venkatesh harugeri  (8 years ago)

Which one is correct answer

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SHATADRU SOBHAN PANIGRAHI  (8 years ago)

Ans is 56 .... let the digis are x, y ,z
then if we ll take 0 ,1,2,3 as one variable then there are 14 posibilties in each 4 case then the total no possibilties are 56

Is this solution Helpfull? Yes | No (1)  

stalin  (8 years ago)

asnwer is 294 ..... and other solved answers below are wrong

Is this solution Helpfull? Yes | No (3)  

Abhijit Das  (8 years ago)

@prashanthi.. the three values can be hold by any of the three variables.. not necessery in the sequence of X Y Z.. they cab be arranged themselves in 3! ways.. so the final answer is -- 110X3!=660...

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