Sorry if I'm wasting everyone's time. I'm checking a homework set, and I'm a little confused by the solution manual's answers, and want a second opinion before shrugging it off as a mistake.

Consider the 25 players on a professional baseball team. At any point, 9 players are on the field.

  1. How many 9-player batting orders are possible given that the order of batting is important?
  2. How many 9-player batting orders are possible given that the all-star designated hitter must be batting in the fourth spot in the order?
  3. How many 9-player fielding teams are possible under the assumption that the location of the players on the field is not important?

I'm getting:

  1. ${25 \choose 9} = 2042975$
  2. ${24 \choose 8} = 735471$
  3. $\frac{25!}{(25-9)!} = 741354768000$

However the solution manual lists:

  1. $7.41\cdot 10^{11}$
  2. $2.97 \cdot 10^{10}$
  3. $2.04 \cdot 10^{6}$

Am I making a dumb, sleep deprived mistake? Thank you for your time.

  • 3
    $\begingroup$ Order is important. For your first question, the answer ${25\choose9}$ is the number of possible sets of $9$ players. This means that (if players are numbers from $1$ to $25$, that the set $\{1,2,\dots,9\}$ will only be counted once. It should be counted more than once as there is more than one ordering of these elements. $\endgroup$
    – 5xum
    Commented Apr 2, 2014 at 8:21
  • 1
    $\begingroup$ Oh my gosh, I'm an idiot. $\endgroup$ Commented Apr 2, 2014 at 8:23
  • 1
    $\begingroup$ I don't know you enough to make judgements of that type, but this mistake does not make you an idiot. It just means you are sloppy and probably still a beginner. Nothing wrong with that. happens to everyone. $\endgroup$
    – 5xum
    Commented Apr 2, 2014 at 8:27
  • 1
    $\begingroup$ @5xum I think you are reading too much into OP's idiot comment. He was probably just mentally kicking himself. ;) $\endgroup$
    – Guy
    Commented Apr 2, 2014 at 8:38
  • 1
    $\begingroup$ @Sabyasachi I know, I know. I was just trying to console him that he made a normal mistake which can happen to everyone. $\endgroup$
    – 5xum
    Commented Apr 2, 2014 at 8:39

1 Answer 1


I reckon there are two steps to be performed:

Step 1. Select a group of $9$ players to take the field out of a team of $25$, and there are $25 \choose 9$ ways of doing this.

Step 2. The number of ways this step can be performed is different in each of the three cases given in your problems. For example, in case 2, there are $9!$ ways.

Finally, we use the multiplication principle and multiply the answer in steps 1 with that in step 2 to obtain the final answer.

Hence your answers should as follows:

  1. $${ 25 \choose 9} \cdot 9! = \frac{25!}{9! (25 -9)!} \cdot 9! = \frac{25!}{16!}.$$

  2. $${ 24 \choose 8} \cdot 8! $$ since one of the players, the all-star designated hitter, must be in the batting team and must be at spot 4.

  3. $$ 25 \choose 9$$


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