The number of word groups possible by taking at least $4$ letters from $3$ words 
The number of word groups by taking at least $4$ letters of each words length, stroke and number are:
$$\text{(A)}\ 221\quad\text{(B)}\ 1068\quad\text{(C)}\ 22\quad\text{(D)}\ 66$$

My Attempt:
Selecting 4 letters from each words: $^6C_4 \cdot ^6C_4 \cdot ^6C_4 = 15^3 = 3375 $
Selecting 5 letters from each words: $^6C_5 \cdot ^6C_5 \cdot ^6C_5 = 6^3 = 216 $
Selecting all 6 letters from each words: $^6C_6 \cdot ^6C_6 \cdot ^6C_6 = 1 $
Thus total words group are $33375 + 216 + 1 = \boxed{3592}$
I don't know what is wrong. Is there any mistake in understanding the question?
 A: Guessing the intent a little... I'm assuming we have to pick a number of different letters that are in the given words and fulfill the criterion of having at least four letters in common with each source word. Your answer, which I would say is also a valid interpretation, doesn't account for the potential of selecting different numbers of letters from different words (so your answer should be even higher).
I think the point might be to exploit the fact that each word has three unique letters, two that are respectively shared with only one of the other two words, and one letter shared by all three source words:

To fulfill the criterion, we can break down choices relative to what we pick from the shared letters ENTR:

*

*All $4$, ENTR: we must choose at least one more from each word, and up to all three. So $(2^3-1)^3 = 7^3 = \fbox{343}$ options for this (meaning the only valid answer choice left would be $1068$, but we'll see).

*E plus two of NTR: So we'll have at least one extra letter from one word and at least two from each of the others, so $\binom 32\left(7\cdot 4^2\right) = 3\cdot 112  = \fbox{336}$ options.

*E plus one of NTR: Now we need at least two extra letters from two words and all three unique letters from the third so $\binom 31\left(4^2\cdot 1\right) = 3\cdot 16 = \fbox{48}$ options.

*E only, none of NTR: This needs all three unique letters from all source words, $\fbox{1}$ option only.

for a total of $343+336+48+1 = \fbox{728}$, not an available option...
So, change of plan, let's limit it to getting exactly four letters from each word, in which case:

*

*All $4$, ENTR:  one extra letter per word, $3^3 = \fbox{27}$ options here.

*E plus two of NTR: So exactly one extra letter from one word and two from each of the others, so $\binom 32\left(\binom 31{\binom 32}^2\right) = 3^4 = \fbox{81}$ options.

*E plus one of NTR: two extra letters from two words, all three from the third so $\binom 31\left({\binom 32}^2\binom 33\right) = 3^3 = \fbox{27}$ options.

*E only, none of NTR: again $\fbox{1}$ option only.

Then this total - for choosing a set of letters that has exactly four letters in common with each source word - is also different to any offered, $27+81+27+1 = \fbox{136}$.
So in conclusion: Interesting, but I still don't know what the original answers related to.
