Assuming that we do not care about labels (i.e. -1 DeAza(outfield), -2 Par (d.h.), -3 Young (outfield),... is considered to be the same outcome as -1 DeAza(d.h.), -2 Par (outfield), -3 Young (outfield),...), as Andre points out this is a perfect time to use the Addition Principle.
Our lineup will consist of 1 catcher, 3 outfielders, 4 infielders, and one additional person (the designated hitter) from one of those categories or from the pitcher category. Split it up into the four cases:
- A) The designated hitter is a catcher
- B) The designated hitter is an infielder
- C) The designated hitter is an outfielder
- D) The designated hitter is a pitcher
Notice, that there is absolutely no way that when we count scenarios from each of these cases, that we accidentally count the same scenario a second time in a different case. I.e., these events are mutually exclusive and therefore we can apply the addition principle and add the amount of possibilities from each individual case to get the number of possibilities total.
Let us examine what happens in one of the cases in closer detail.
- Case A) The designated hitter is a catcher.
This means that among our nine players in the batting lineup, two of them are catchers, 3 of them are outfielders, and 4 of them are infielders. We wish to count how many different lineups satisfy this statement. Here we will use the Multiplication Principle. Set up a sequence of questions/choices to determine which lineup we have.
- Pick which two of the catchers are in the lineup ($\binom{2}{2}=1$ choice since there are two catchers available and we want to choose two of them to appear in our lineup)
- Pick which three of the outfielders are in the lineup ($\binom{6}{3} = 20$ choices since there are six outfielders available and we want to choose three of them to appear in our lineup)
- Pick which four of the infielders are in the lineup ($\binom{5}{4} = 5$ choices since there are five infielders available and we want to choose four of them to appear in our lineup)
- Pick which zero of the pitchers are in the lineup ($\binom{12}{0} = 1$ choice since there are twelve pitchers available and we don't want to choose any of them, i.e. the empty choice)
- Pick which order the nine people chosen appear in the lineup ($9! = 362880$ choices since every way of ordering the players counts as "different")
Thus, by the multiplication principle there are $\binom{2}{2}\binom{6}{3}\binom{5}{4}\binom{12}{0}9!=36288000$ different outcomes for the first case (case A).
Calculate the number in case B, case C, and case D similarly (in your comment on the other copy you only used a $\binom{12}{1}$ for case D, however that only counts how many ways to pick one pitcher to be the designated hitter; you still need to pick the rest of the team and where they are in the lineup).
For part B, we can approach this two ways. The easier way I think is this: first ask the question "What is the probability that Jones is not picked to be in the starting lineup?" By answering this question, we are able to then easily answer the follow up question "What is the probability that Jones is in the starting lineup?".
We will answer the question of the probability that he is not exactly like how we answered for the first part of the question with the following adjustment: since we want to count the number of cases Jones is not in the lineup, redo the calculations for the first part as though Jones weren't on the team at all (I.e. remove Jones from the list of available outfielders, making it so that there are only $5$ outfielders available).
We again break it into the same cases as before: A) d.h. is a catcher, B) d.h. is an infielder, C) d.h. is an outfielder, D) d.h. is a pitcher.
Let us examine the first case in detail again:
- Pick which two of the catchers are in the lineup ($\binom{2}{2}=1$ choice since there are two catchers available and we want to choose two of them to appear in our lineup)
- Pick which three of the outfielders are in the lineup ($\binom{5}{3} = 10$ choices since there are $\color{red}{\star}$five$\color{red}{\star}$ outfielders available and we want to choose three of them to appear in our lineup) (since we don't want Jones)
- Pick which four of the infielders are in the lineup ($\binom{5}{4} = 5$ choices since there are five infielders available and we want to choose four of them to appear in our lineup)
- Pick which zero of the pitchers are in the lineup ($\binom{12}{0} = 1$ choice since there are twelve pitchers available and we don't want to choose any of them, i.e. the empty choice)
- Pick which order the nine people chosen appear in the lineup ($9! = 362880$ choices since every way of ordering the players counts as "different")
For a total of $\binom{2}{2}\binom{5}{3}\binom{5}{4}\binom{12}{0}9!$ different outcomes for the first case. Calculate the other cases similarly.
Now that we have a tally of how many lineups do not have Jones on the roster, take the answer from the first half of the problem and subtract the number just calculated here to get the number of different ways that Jones IS on the starting lineup (as per a simple application of inclusion-exclusion ).
Take this newest number and divide by the answer found in the first half to get the probability that Jones is in the starting lineup as per the definition of probability in an equiprobable sample space:
$$Pr(A) := \frac{|A|}{|S|}$$
Complete the problem by noting that, given that Jones is in the starting lineup, he is equally likely to be in each of the spots and apply the multiplication rule (probability).
$$Pr(\text{Jones bats}~4^{th}) = Pr(\text{Jones in lineup})\cdot Pr(\text{Jones bats}~4^{th}~|~\text{Jones in lineup})$$
There is another equally acceptable approach to solving the second part. Directly count the number of ways that Jones bats fourth. Do so by breaking it into cases:
- A) The designated hitter is a catcher
- B) The designated hitter is an infielder
- C) The designated hitter is a pitcher
- D) The designated hitter is an outfielder
Looking at one of these cases more closely, looking at A again: Break it up via multiplication principle as before. Since the D.H. is a catcher, that means our lineup will consist of 2 catchers, Jones, 2 other outfielders, and 4 infielders.
- Pick which two of the catchers are in the lineup ($\binom{2}{2}=1$ choice since there are two catchers available and we want to choose two of them to appear in our lineup)
- Pick which two of the other outfielders are in the lineup ($\binom{5}{2} = 10$ choices since there are $\color{red}{\star}$five$\color{red}{\star}$ outfielders available and we want to choose two of them to appear in our lineup) (since we already know Jones is in the lineup)
- Pick which four of the infielders are in the lineup ($\binom{5}{4} = 5$ choices since there are five infielders available and we want to choose four of them to appear in our lineup)
- Pick which zero of the pitchers are in the lineup ($\binom{12}{0} = 1$ choice since there are twelve pitchers available and we don't want to choose any of them, i.e. the empty choice)
- Pick which order the nine people chosen appear in the lineup ($\color{red}{\star}8! = 40320\color{red}{\star}$ choices since Jones is already occupying the fourth slot, and we want to arrange the other eight players around him)
Do so similarly for the other cases to find the number of ways that Jones appears in the fourth batting position. Complete by applying the definition of probability (num outcomes for event / num outcomes regardless).
Best Answer
Offhand this sounds like an integer programming problem, but you haven't really specified what it is you're looking to optimize. That is, if we list the available players as $i=1,2,3,\ldots,N(=154)$, let $p_i$ be player $i$'s draft price, and let $x_i=1$ or $0$ indicate whether you draft player $i$ or not, then two constraints are
$$\sum_i x_i=14\quad\text{and}\quad\sum_ip_ix_i\le C$$
where $C$ is the salary cap. There will be additional constraints corresponding to the requirements on the maximum or minimum number of catchers, outfielders, etc. But the question is, how do you describe mathematically the players' combined "contributions"?
For example, suppose we limit things to two categories, say home runs and stolen bases, and suppose player $i$'s value in each category is $h_i$ and $s_i$, respectively. We might define
$$H(x_1,\ldots,x_N)=\sum_i h_ix_i\quad\text{and}\quad S(x_1,\ldots,x_N)=\sum_is_ix_i$$
But we would still need to specify an objective function $W(H,S)$ to optimize, and it sounds like that function, whatever it is, is going to be nonlinear (since you indicated there are diminishing returns for each category). In short, you have not yet posed your question here in a way that can be answered mathematically.