Forming a Commitee of Senators From Blitzstein & Hwang "Introduction to Probability":

The US senate consists of 100 senators, with 2 from each of the 50 states. There are $d$ Democrats in the Senate. A committee of size $c$ is formed, by picking a random set of senators such that all sets of size $c$ are equally likely. Find the expected number of Democrats on the committee.

Attempts at solving: 
i) Assumed that there were ${100}\choose{c}$ total combinations and then tried $X\sim Binom(100,\cfrac{d}{100})$, where $X$ is the number of Democrats in the commitee however then realised that if we have two independent variables ($c$ and $d$), then there will be some cases that will be nonsensical; if we fix $c=10$ and $d=5$, then we cannot fill up the committee with Democrats so we will have overcounted.
ii) Attempted the problem in a similar fashion to (i), however tried fitting $X$ to: $X \sim HGeom(d,100-d,100)$ however ended up at the same issue of overcounting as a result of two independent variables again.


After this I concluded that $X$ would not be distributed as a named distribution but I am not having any luck in dealing with the issue of two independent variables (some pointers would be much appreciated).
 A: The fraction of democrats in the senate is $\frac{d}{100}$. The expectation for a random committee of size $c$ would be the same fraction.  So the expected number of democrats on the committee would be $\frac{d}{100}\cdot c=\frac{cd}{100}$.
This problem can also be approached by choosing senators sequentially for the committee.  Let $X_i$ be $1$ if the $i$-th senator is a democrat, and $0$ if not.  The number of democrats is $X_1 +\cdots +X_c$.  Then the expected number of democrats is $E[X_1+\cdots+X_c]=E[X_1]+\cdots+E[X_c]$ (by linearity of expectation, which does not require independence).  Each $E[X_i]$ is $\frac{d}{100}$, from which the same result follows.
A: [The binomial distribution has the effect of sampling with replacement, so you don't want that part.]
The number of equally likely ways in which to choose a committee of size $c$
is ${100 \choose c}$ as you say.
The number of ways to choose such a committee with $D$ Democrats and
$c - D\,$ non-Democrats is ${d \choose D}{100 - d \choose c-D}.$ In order
to make sense of this, we must have $c \ge D.$
This means that the random variable $D$ that counts the Democrats on
a randomly selected committee of size $c$ is hypergeometric. The hypergeometric
mean is $E(D) = c(d/100).$ 
Notice that it is necessary to distinguish between the number $d$ of 
Democrats in the senate, and the random number $D$ of Democrats on the
committee.
Given the text you are using (thanks for telling us), you will probably
be used to the idea of verifying analytic results via simulation.
Here are a million simulated committees of size $c=14$ in a (hypothetical) Senate with 42 Democrats.
 m = 10^6 # iterations
 D = numeric(m)
 c = 14   # committee members
 d = 42   # Democrats in Senate
 sen = c(rep(1,d), rep(0, 100-d)) # 1 = Dem, 0 = non-Dem
 for(i in 1:m) {
   com = sample(sen, c)  # randomly select committee
   D[i] = sum(com) }
 mean(D)  # aprx E(D)
 ## 5.880656
 c*(d/100) # exact E(D)
 ## 5.88
 sd(D)    # aprx SD(D)
 ## 1.720396
 2*sd(D)/sqrt(m)
 ## 0.003440792

With a million iterations $m$, the simulation
estimates $E(D) \approx 5.881 \pm 0.003,$ compared with the exact
result $E(D) = 5.88.$  According to the Central Limit Theorem, the 95% margin
of simulation error is $1.96\,SD(D)/\sqrt{m}.$ Your text should have
a formula for $Var(D)$ of a hypergeometric random variable $D,$ so you can check the accuracy of the simulated standard
deviation for yourself.
Below is a histogram of the simulated distribution of $D$, where red dots
atop histogram bars show exact hypergeometric probabilities. The vertical
resolution of the graph is about 0.002, so the fit looks almost perfect.
 hist(D, prob=T, br=(0:15)-.5, col="skyblue")
   i = 0:15;  pdf = dhyper(i, d, 100-d, c)
   points(i, pdf, pch=19, col="red")


