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.
