In how many ways can you select one of the two but not both? For this question:
A committee of three boys and three girls is to be selected from a class of 14 boys and 17 girls. In how many ways can the committe be selected if:
a.) Ana has to be on the committee.
Ana is a girl so:
$$(^{14}C_3)(^{16}C_2)=43680$$
b.) the girls must include either Roberta or Priya, but not both.
I basically needed help in this question (Question B)
Attempts I made so far:
1.) $$(^{14}C_3)*2(^{16}C_2)-(^{15}C_1)=87345$$
2.) $$(^{14}C_3)*2[(^{16}C_2)-(^{15}C_1)]=76440$$
I need help in what I've been doing wrong in these workings for question B
 A: What you want to use is something called the Principle of Inclusion/Exclusion. Consider first the committees that Roberta is on. We include all such committees, giving $\binom{14}{3}\binom{16}{2}$. But we don't want all of these, since Priya could be on some. So let's exclude the committees in which both Roberta and Priya are on. There are $\binom{14}{3}\binom{15}{1}$ such committees, so there are $\binom{14}{3}\binom{16}{2}-\binom{14}{3}\binom{15}{1}$ committees that contain Roberta but not Priya. The same process shows that there are $\binom{14}{3}\binom{16}{2}-\binom{14}{3}\binom{15}{1}$ committees that contain Priya but not Roberta. Thus we add the two cases, giving $2\left(\binom{14}{3}\binom{16}{2}-\binom{14}{3}\binom{15}{1}\right)$ total committees. 
A: Let's focus on the number of ways the girls can be selected.  Let event $P$ be the event that Priya is selected to be one of the girls on the committee.  Let event $R$ be the event that Roberta is selected to be one of the girls on the committee.  The Inclusion-Exclusion Principle states that the number of committees that contain Priya or Roberta is 
$$|P \cup R| = |P| + |R| - |P \cap R|$$
where $|S|$ denotes the number of elements in set $S$.
The number of ways we can select the girls for the committees that contain Priya is $\binom{16}{2}$ since we must select Priya and two of the remaining sixteen girls.  The number of ways we can select the girls for the committees that contain Roberta is also $\binom{16}{2}$ since we must select Roberta and two of the remaining sixteen girls.  The number of ways we can select the girls for the committees that contain both Priya and Roberta is $\binom{15}{1}$ since we must select Priya, Roberta, and one of the remaining fifteen girls.  Thus, the number of ways we can select the girls for the committees that contain Priya or Roberta is 
$$|P \cup R| = |P| + |R| - |P \cap R| = \binom{16}{2} + \binom{16}{2} - \binom{15}{1} = 2\binom{16}{2} - \binom{15}{1}$$
However, we must exclude those selections in which we select both Priya and Roberta.  Therefore, the number of ways we can select the girls for the committees that contain Priya or Roberta, but not both, is 
\begin{align*}
|P \triangle R| & = |P \cup R| - |P \cap R|\\
                & = 2\binom{16}{2} - \binom{15}{1} - \binom{15}{1}\\
                & = 2\binom{16}{2} - 2\binom{15}{1}\\
                & = 2\left[\binom{16}{2} - \binom{15}{1}\right]
\end{align*}
For each such selection, we can choose the boys in $\binom{14}{3}$ ways.  Therefore, the number of committees that can be selected that contain Priya or Roberta, but not both, is 
$$\binom{14}{3} \cdot 2\left[\binom{16}{2} - \binom{15}{1}\right]$$
The symmetric difference of sets $A$ and $B$, denoted $A \triangle B$, is the set of elements that are in exactly one of the sets $A$ or $B$.
$$A \triangle B = (A \cup B) - (A \cap B) = (A - B) \cup (B - A)$$
I computed the result by subtracting the number of elements in the intersection from the number of elements in the union.  @bburGsamohT arrived at the same result by adding the number of elements in each set difference.  
A: How about ${2\choose 1}{14\choose 3}{15\choose 2}=76440$
i.e. we select one of the two girls with 3 from 14 boys and then we have to select the remaining 2 girls from 15
