Number of ways to schedule tennis matches against five players on 10 different days Conditions:

*

*You play one tennis match a day and have 10 days available to play

*You play each player twice

*One player cannot play on day one

How many different ways can you schedule these matches?
My thought process:

*

*Day 1: 4 ways (one cannot play)

*Day 2 - 10: 9 * 8c4 ways (4 players are playing twice 8c4 ways, 1 plays once 9 different ways)

*total ways = 4 * 9 * 70 = 2,520 ways
Does this approach seem right?
Thanks!
 A: 
Does this approach seem right?

Hmmm,... actually its fairly close.

"4 players are playing twice 8c4 ways, "

Rather, there are $8!/2!^4$ ways to assign 2 days from 8 to each of 4 players.  It is a multinomial selection.
Solution: We calculate the number of ways to select two days for each player, with one player only available for nine days (so count ways to select days for that player first).
$$\dfrac{9!}{2!7!}\cdot \dfrac{8!}{2!^4}=90720$$
A: Suppose we look at how many ways to assign the players if all the players could play on any given day.  In this case, we have $_{10}C_2$ ways of assigning days to Player 1, $_8C_2$ ways of subsequently assigning days to Player 2, $_6C_2$ for Player 3, $_4C_2$ for Player 4, and $_2C_2$ for Player 5.  Multiplying these together gives $\frac{10!\cdot8!\cdot6!\cdot4!\cdot2!}{(2!\cdot8!)(2!\cdot6!)(2!\cdot4!)(2!\cdot2!)(2!\cdot0!)}=\frac{10!}{(2!)^5}=113400$ ways of assigning 5 players matches on 2 days each.
Suppose Player 5 cannot play on Day 1.  In $\frac15$ of these ways, Player 5 was assigned to Day 1.  Therefore, we need to reduce our number of ways by multiplying by $\frac45$, and we have a final total of 90720
