So we have 4, 6 and 10 cents and we can use that to make $n$ cents. In how many ways can we do that if order doesn't matter?
so I tried to solve it and got $T(n) = T(n-10) + T(n-6) + T(n-4) - c$ it works for the case when order matters and $c=0$ i tried changing the value of $c$ to eliminate the order but depending on the value of $c$ it works for some numbers and not others. What do I change so that order doesn't matter. Thanks.
One approach is to say that the generating function is $$\frac{1}{(1-x^4)(1-x^6)(1-x^{10})} = \frac{1}{1-x^4-x^6+ x^{14}+ x^{16}-x^{20} }$$ and so a recurrence is $$T(n)=T(n-4)+T(n-6)-T(n-14)-T(n-16)+T(n-20)$$ starting with $T(0)=1$ and $T(n)=0$ when $n <0$
Without generating functions then, by considering what happens if you add a single coin of a particular value and using earlier results,
you can say that the number of ways of making $n$ cents just with $4$-cent coins satisfies $$F(n)=F(n-4)$$
while the number of ways of making $n$ cents with $4$- and $6$-cent coins satisfies $$S(n)=S(n-6)+F(n) = S(n-6)+F(n-4) = S(n-6)+S(n-4)-S(n-10)$$
so the number of ways of ways making $n$ cents with $4$- and $6$- and $10$-cent coins satisfies $$T(n)=T(n-10)+S(n) = T(n-10)+S(n-6)+S(n-4)-S(n-10)$$ $$= T(n-10)+T(n-6)-T(n-16)+T(n-4) -T(n-14)-T(n-10)+T(n-20)$$
and if you tidy this up then you get the same recurrence as with generating functions