Question 47

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Jambmaths question: 

$\begin{align}  & \text{In how many ways can a students select  2 subjects from 5 subjects} \\ & \text{(A) }\frac{5!}{2!}\text{  (B) }\frac{5!}{3!}\text{  (C) }\frac{5!}{2!2!}\text{  (D) }\frac{5!}{2!3!} \\\end{align}$

Jamb Maths Solution: 

$\begin{align}  & \text{The problem deals with selection(i}\text{.e combination)} \\ & \text{Possible selection is given by}{{\text{ }}^{n}}{{C}_{r}}=\frac{n!}{r!(n-r)!} \\ & \text{Therefore selecting 2 subjects from 5 subjects will} \\ & \text{be}{{\text{ }}^{5}}{{C}_{2}}=\frac{5!}{2!(5-2)!}=\frac{5!}{2!3!} \\\end{align}$

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