From knowing
$7c=\sum_{i=1}^{50-c}k_i$
and
$c\choose 2 $=$ \sum_{i=1}^{50-c}$ $k_i\choose 2 $
how can I get to
$\sum_{i=1}^{50-c}(k_i-\mu)^2=(50-c)\mu^2-14c\mu+c^2+6c$
for some arbitrary $\mu$?
I would show you the many attempts I have made at this but I am sure it will look like gibberish. This is not homework.
Well, I did all that ugly derivations for you.
Let $n = 50 - c$.
Assume:
$$\sum_{i=1}^{n} k_i = 7c$$
$${c\choose 2}=\sum_{i=1}^{n} {k_i\choose 2}$$
Obtain:
$$\sum_{i=1}^{n} (k_i - \mu)^2 = n\mu^2 - 14c\mu + c^2 + 6c$$
Proof.
(It looks like a variance.)
$$(k_i - \mu)^2 = k_i^2 - 2k_i\mu + \mu^2$$
$$\sum_{i=1}^n (k_i - \mu)^2 = \sum_{i=1}^n k_i^2 - 2k_i\mu + \mu^2$$ $$\sum_{i=1}^n (k_i - \mu)^2 = \sum_{i=1}^n k_i^2 - 2\mu \sum_{i=1}^n k_i + \mu^2 \sum_{i=1}^n 1$$ $$\sum_{i=1}^n (k_i - \mu)^2 = \sum_{i=1}^n k_i^2 - 2\mu \sum_{i=1}^n k_i + \mu^2 n$$
Then
$${k_i\choose 2} = \frac {(k_i - 1) k_i} 2$$
$$\sum_{i=1}^{n} {k_i\choose 2} = \frac 1 2 \left(\sum_{i=1}^{n} k_i^2 - \sum_{i=1}^{n} k_i\right) = \frac {(c - 1)c} 2$$
$$\sum_{i=1}^{n} k_i^2 - \sum_{i=1}^{n} k_i = (c - 1)c$$
And substitute it all together
$$\sum_{i=1}^n (k_i - \mu)^2 = (c - 1)c + \sum_{i=1}^{n} k_i - 2\mu \sum_{i=1}^n k_i + \mu^2 n$$
$$\sum_{i=1}^n (k_i - \mu)^2 = (c - 1)c + 7c - 2\mu 7c + \mu^2 n$$