Three variables x y z have sum of 30 All three of them are non-negetive integers If any two variables

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
8 years agoHelpfull: Yes(5) No(10)
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
8 years agoHelpfull: Yes(1) No(3)
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
8 years agoHelpfull: Yes(0) No(4)
which one is correct plzzztell me?//
8 years agoHelpfull: Yes(0) No(3)
Which one is correct answer
8 years agoHelpfull: Yes(0) No(1)
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
8 years agoHelpfull: Yes(0) No(1)
asnwer is 294 ..... and other solved answers below are wrong
8 years agoHelpfull: Yes(0) No(3)
@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...
8 years agoHelpfull: Yes(0) No(4)

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.