There is a group of $5$ children, where two of the children are twins. How many ways can I distribute $6$ identical pieces of candy to the children, if the twins must get an equal amount of candy?

50

Step-by-step explanation:

Suppose that the kids get $x_1,x_2,x_3,x_4,x_4$ pieces of candy, respectively (each twin received $x_4$ pieces). Then we seek all quadruples $(x_1,x_2,x_3,x_4)$ satisfying

\[x_1+x_2+x_3+2x_4 = 6.\]We proceed by casework, based on the number of candies the twins receive.

If $x_4 = 0$, then there are 6 candies to distribute to 3 kids. This is like arranging 6 C's and 2 |'s (dividers), so there are $\binom{8}{2} = 28$ possible distributions.

If $x_4 = 1$, there are 4 candies remaining for the other 3 kids, so there are $\binom{6}{2} = 15$ possible distributions.

If $x_4 = 2$, there are 2 candies remaining, so there are $\binom{4}{2} = 6$ possible distributions.

If $x_4 = 3$, there are 0 candies remaining, for 1 possible distribution.

Adding all of these results from the separate cases, there are a total of

\[28 + 15 + 6 + 1 = \boxed{50}\].

only 10/60 isn't equivalent

hope this : )