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
Best Answer
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.