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Maths Puzzle
Numerical Ability
Number System
A three digit number of the form ‘xyz’ is to be formed such that x > y, x > z, and y < z. How many such numbers are possible if x > 0?
Read Solution (Total 1)
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- x>z>y
x,y,z can be 0,1,2,.....9
x=0 is not possible as x>0.
so, if x=1, then z=0 which in turn gives y=0 which is not possible as z>y not z=y.
the valid cases starts from x=2 where z=1,y=0.
next case is x=3 and z=2 or 1 and y=1 or 0.
following the same procedure again and again....
..........................................................
the last case is when x=9 which provides z with 8 choices and similarlu provides y with 8 choices.
so according to the above deduction it forms a standard result as:
1+4+9+16+.........+64.
=204 - 8 years agoHelpfull: Yes(0) No(0)
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