Counting possibilities There are 4 different sports which are to be played on a seven day program such that on each day exactly one sport can be played. If Mr D must play at least two sports on at at least three days each, in how many ways can he schedule the program? 
My approach is as follows
Let the two sports he plays on 3 days are $X$ and $Y$. 
For remaining one day, he can either play:
Case $1$: One of the $X$ or $Y$
OR 
Case $2$: Another sport $Z$
Case $1$:
$^7C_4 \times 4 \times 3$ 
choose $4$ days out of $7$ on which $X$ will be played. Multiply by $4$ as number of ways of choosing $X$ is $4$. Multiply by $3$ as $Y$ can now be chosen in $3$ ways.
Total $ 420$
Case 2: 
$^7C_3 \times 4 \times ^4C_3 \times 3 \times 2$
choose $3$ days out of $7$ on which $X$ is to be played. Multiply by number of possibilities of $X$ i.e $4$. Choose $3$ days out of remaining $4$ days on which $Y$ is to be played and multiply it by number of possibilities of $Y$ i.e $3$. Multiply by $2$ i.e. the number of possibilities for $Z$ which shall be played on the remaining one day. 
Total $3360$
Answer key shows Case $2$ totals $1680$ ways, so my approach has to be wrong. Can't figure out where. 
 A: You counted each possible schedule in Case 2 twice.  To see this, observe that choosing to play sport $X$ on Monday, Wednesday, and Friday and then choosing to play sport $Y$ on Tuesday, Thursday, and Saturday has the same effect as first choosing to play sport $Y$ on Tuesday, Thursday, and Saturday and then choosing to play sport $X$ on Monday, Wednesday, and Friday.  Therefore, it does not matter in which the two sports that are played on three days are selected.
Choose which two of the four sports are played on three days each.  Of these two, choose on which three of the seven days the sport that appears first on an alphabetical list is played.  Next, choose on which three of the four remaining days the other sport that will be played on three days is played.  Finally, choose which of the other two sports will be played on the remaining day.
$$\binom{4}{2}\binom{7}{3}\binom{4}{3}\binom{2}{1}$$
Notice that 
$$\frac{1}{2}\binom{7}{3} \cdot 4 \cdot \binom{4}{3} \cdot 3 \cdot 2 = \binom{4}{2}\binom{7}{3}\binom{4}{3}\binom{2}{1}$$
