Answer to Question #98475 in Discrete Mathematics for Ahmed

Question

In an exam, a student is required to answer 10 out of 13 questions. Find the number of
possible choices if the student must answer:
(a) the first two questions;
(b) the first or second question, but not both;
(c) exactly 3 out of the first 5 questions;
(d) at least 3 out of the first 5 questions.

Answer 1

a)

The student MUST answer to the first 2 questions. These two questions are not fit to be chosen. So we have 13-2=11 questions left, and 10-2=8 questions left to be chosen.

$$n=\binom{13-2}{10-2}={11! \over 8!(11-8)!}={11(10)(9) \over 1(2)(3)}=165$$

b)

One question, the first or the second, has been answered. So we have 13-1=12 questions left, and we have to answer to 10-1=9 questions.

$$n=\binom{2}{1}\binom{13-2}{10-1}=2\cdot{11! \over 9!(11-9)!}=2\cdot{11(10) \over 1(2)}=110$$

c)

$$n=\binom{5}{3}\binom{13-5}{10-3}={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}=10\cdot8=80$$

d)

$$n=\binom{5}{3}\binom{13-5}{10-3}+\binom{5}{4}\binom{13-5}{10-4}+\binom{5}{5}\binom{13-5}{10-5}$$

$$={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}+{5! \over 4!(5-4)!}\cdot{8! \over 6!(8-6)!}+{5! \over 5!(5-5)!}\cdot{8! \over 5!(8-5)!}$$

$$=10(8)+5(28)+1(56)=276$$