Given, f(1)+f(2)=f(3)
Now possible values of f(3) are 2,3,4,5 and 6 (it cannot be 1 as minimum value f(1)&f(2) is 1, so 1+1=2 )
For each k, if f(3)=k then there will be (k−1) set of values for f(1) and f(2)
For example if f(3)=3 then possible value of f(1)&f(2) will be (1,2)or(2,1)⇒ 2 values
Similarly for f(3)=4, possible values will be (1,3),(3,1),(2,2)⇒3 values.
And so on.
So number of possible combinations of f(1),f(2) and f(3) are 1+2+3+4+5=15
And for each value of f(3) there will be 6 possible value of f(4), so total cases will be 15×6=90