Answer to Question #262367 in Discrete Mathematics for zid

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)

$\dbinom{13}{10}=\dfrac{13!}{10!(13-10)!}=\dfrac{13(12)(11)}{1(2)(3)}=282$

(b)

$\dbinom{2}{1}\dbinom{11}{9}=2\cdot\dfrac{11!}{9!(11-9)!}=\dfrac{2(11)(10)}{1(2)}=110$

(c)

$\dbinom{5}{3}=\dfrac{5!}{3!(5-3)!}=\dfrac{5(4)}{1(2)}=10$

(d)

$\dbinom{5}{3}+\dbinom{5}{4}+\dbinom{5}{5}=10+5+1=16$