I'll get by with a little help ....

Because the students are selected randomly, all have the same chance of being selected to go on the field trip, so it doesn't really matter how they are selected. So the answer will be the same as if you picked $25$ from $50$ in any other fair way.

So: The probability that the first of the friends is picked to go is $\frac{25}{50}$ , the probability that the second is $\frac{24}{49}$ and the third: $\frac{23}{48}$ .

The probability that all three go is

$\frac {25 \times 24 \times 23}{50 \times 49 \times 48} = \boxed{\frac{23}{196}}$

The probability is

$(\dfrac{1}{8}) * \dfrac{47*46*45*43*41* ..... *3*1}{49*47*45*43*41* .... *3*1} = \dfrac{23}{196}$ ,

giving us $a = 23, b = 196$ and $a + b = \boxed{219}$ .

I'll post my method in the morning, but I just wanted to post something now so that others could make comments and inquiries.

Edit: O.k., now for an explanation of the above calculation.

First, we need each of the $3$ friends to be in different pairings, for if two are paired together then it is certain that one of them will not be going on the field trip. So we need to calculate the probability that all $3$ are in separate pairings.

Now choose one of the $3$ , say Silas. Then we can pair him up with one of the $47$ remaining students. We can then pair Suyeon up with one of the remaining $46$ , and then Calvin with one of the remaining $45$ students. This leaves us with $44$ students left to pair up. Choose one of these at random and pair them up with one of the remaining $43$ students. Now we're left with $42$ students. Choose one at random and pair them up with one of the remaining $41$ students. Continue this process until the last two students are paired by default. This gives us $47*46*45*43*41* ..... *3*1$ ways to pair up the students so that none of the three friends are paired together.

Next, without restrictions, we can count the number of possible pairing arrangements in the same way we did the remaining $44$ students in the previous paragraph. This gives us $49*47*45*43*41* ..... *3*1$ possible pairing arrangements.

Thus the probability that a pairing arrangement has the three close friends separated is $\frac{46}{49}$ . But now we need each of the three friends to win their pair's coin toss. Assuming the coin tosses are fair, all three will win with a probability of $(\frac{1}{2})^{3} = \frac{1}{8}$ . So the desired probability is $(\frac{1}{8})(\frac{46}{49}) = \frac{23}{196}$ .

Obviously, any two of the three friends cannot be sorted under the same group. $P=\frac{47}{49} \times \frac{46}{47}=\frac{46}{49}$ The first fraction means the probability of the first one of the friends having his or her partner who is not any other two of the friends.

The second fraction means under above condition, the second one of the friends having his or her partner who is not the remaining one of the friends.

Now students are paired up with the three friends each in different groups.

There is further selection of flipping coins which has probability of $\frac{1}{2}$ each.

$P=\frac{46}{49} \times (\frac{1}{2})^{3}=\frac{23}{196}$

Hence, a+b=23+196= $\boxed{219}$

Since the probability requirement is that we need the specifically choose the 3 students to go to the field trip out of the 50 students, hence, we only need to choose 22 out of the remaining 47 students . Hence, the probability may be expressed as :

$\dfrac{47 \choose 22}{50 \choose 25}$ $= \boxed{\dfrac{23}{196}}$

Thus, $a + b = 23 + 196 = 219$