**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.**

The **Answer to the Question**

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**Here's the Solution to this Question**

(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)

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