I can't figure out a sensible answer to the b-part and c-parts. Correct answers weren't provided in the book either. I'm self studying Grimaldi's discrete math book.
A) there's 10 people, and we have to select committee of four, with president, vp, secretary, and treasurer.
Order matters it seems.
B) IF the 10 people contain 3 doctors, how many of the original selections have doctor as president?
C) if there's 3 doctors among 10 people, how many selections have exactly one doctor inside?
D) if there's 3 doctors among 10 people, how many selections contain at least 1 doctor?
answers
a) break up selection into choose president in 10 ways, choose vp in 9 ways, choose sec in 8 ways, choose tres in 7 ways. = 5040 ways to select the committee from 10 distinct individuals.
b)assuming that the multiplication principle of counting still works. I would imagine that we first choose president 3 choices from doctors (daniel the doc, donny the doc, debra the doc), second we can actually choose the vp from all the remaining people, because if the president is doctor, then the rest may or may not be doctors, all is allowed except we cannot run out of the doctors. so we would have 3*9*8*7=1512 selections which have doctor as president.
c) exactly 1 doctor inside the selection.
we could start from the basic situation of trying to get the president to be doctor, and others have to be regulars. 3*7*6*5 = 630 HOWEVER the problem is that the doctor can be located anywhere (doctor can be pres, vp, sec or tres).I'm not sure if we have to have different orderings and multiply by four or leave the result at 630.
d) at least 1 doctor inside (1,2, or 3 docs).
I think we could calculate 7*6*5*4 = 840 ways to select committee without any doctors inside. Then we have 5040-840 = 4200 selections with at least 1 doctor inside. I could be wrong here, though...
Your answers to questions A, B and D are correct. As for C, this problem can be solved as follows. First select one doctor, then select three other people. Finally, assign each of the positions to one person. The number of possible assignments thus equals:
$${3 \choose 1}{7 \choose 3}4! = 3 \cdot 35 \cdot 24 = 2520$$
This approach can also be used to calculate the result for question D, first selecting one, then two and finally three doctors:
$$\left({3 \choose 1}{7 \choose 3} + {3 \choose 2}{7 \choose 2} + {3 \choose 3}{7 \choose 1}\right) 4! = \left(3 \cdot 35 + 3 \cdot 21 + 1 \cdot 7\right) \cdot 24 = 175 \cdot 24 = 4200$$