For the redox reaction, `MnO_4^(-) + C_2O_4^(2-) + H^(+) -> Mn^(2+) + CO_2 +H_2O` The correct coefficients of the reactants for the balanced reaction are respectively, `MnO_4^-` , `C_2O_4^(2-)` , `H^+` ? :-

Permanganate, `Mn(VII)`, is REDUCED to `Mn^(2+)`:

`MnO_4^(-) +8H^+ + 5e^(-) rarr Mn^(2+)+4H_2O(l)` `(i)`

Oxalate anion, `C(+III)`, is oxidized to `CO_2`:

`C_2O_4^(2-)rarr2CO_2+2e^(-)` `(ii)`

And we take `2xx(i)+5xx(ii)`

`2MnO_4^(-) +16H^+ +10e^(-)+5C_2O_4^(2-)rarr10CO_2+10e^(-)+5Mn^(2+)+8H_2O(l)`

..to give finally....

`underbrace(2MnO_4^(-))_"bright purple" +5C_2O_4^(2-)+16H^+rarrunderbrace(2Mn^(2+))_"colourless"+10CO_2+8H_2O(l)`

An examiner is being lazy if they set a question like this....

at first you must find the elements that change their oxidation number end calculate these numbers before and after the reaction:

Mn goes from +7 to +2 and it gain 5 electrons:
`MnO_4^(-) + 5 e + 8 H^+= Mn^(2+) + 4 H_2O`
look that you must balance the atoms of oxygen with water and the atoms of Hydrogen with ion`H^+`

while C goes from + 3 to +4 and it lost 1 electron for each atoms
`C_2O_4^(2-) = 2CO_2 + 2 e`

but you must balance the number of electrons in the two reactions, so you mustmoltiplicate for 2 the first and for 5 the second reaction:
`2MnO_4^(-) + 10 e + 16 H^+= 2Mn^(2+) + 8 H_2O`
`5C_2O_4^(2-) =10CO_2 + 10 e`

now you summ the two reactions and you have:
`2MnO_4^(-) + 10 e + 16 H^+5C_2O_4^(2-) =10CO_2 + 10 e+ 2Mn^(2+) + 8 H_2O`

semplifying the electrons, you get
`2MnO_4^(-) + 16 H^+5C_2O_4^(2-) =10CO_2 + 2Mn^(2+) + 8 H_2O`
that is what you were looking for

Because of the `H_2` on the right, the number of `H^+` ions on the left must be even. (4) is eliminated.

Looking at the equation, there must be a significantly smaller number of `MnO_4` ions than `H^+` ions (otherwise there would be far too many oxygen atoms for the carbon and hydrogen atoms to balance them out in the form of `CO_2` and `H_2O`), so (1) is the only choice left.

One can definitely try to balance the equation, but in this case it is unnecessary.