KVPY SA Exam 2018 Question Paper

Digits are 1, 2, 3, 4, 5
$\begin{array}{ccc}\mathrm{Even}\mathrm{digits}=2,4& ;& \mathrm{number}\mathrm{of}\mathrm{Even}\mathrm{digits}=2\\ \mathrm{Total}\mathrm{marks}=2375& ;& \mathrm{number}\mathrm{of}\mathrm{Odd}\mathrm{digits}=3\end{array}$
Sum of any 2 adjacent digits is odd
⇒ Odd and even digits will alternate
Case I For n; Repetition is not allowed
⇒ OEOEO is the only possibility of arrangement of digits, Where O = Odd digit, E = Even digit.
So number of Arrangements $\mathit{n}=\frac{3}{\mathit{O}}\times \frac{2}{\mathit{E}}\times \frac{2}{\mathit{O}}\times \frac{1}{\mathit{E}}\times \frac{1}{\mathit{O}}=12$
Case I For m ; Repetition is allowed
⇒ Two possipilities
(a) OEOEO
Number of such arrangements = $\frac{3}{\mathit{O}}\times \frac{2}{\mathit{E}}\times \frac{3}{\mathit{O}}\times \frac{2}{\mathit{E}}\times \frac{3}{\mathit{O}}=108$
(b) EOEOE
Number of such arrangements = $\frac{2}{\mathit{E}}\times \frac{3}{\mathit{O}}\times \frac{2}{\mathit{E}}\times \frac{3}{\mathit{O}}\times \frac{2}{\mathit{E}}=72$
So $\mathit{m}=108+72=180$
$\frac{\mathit{m}}{\mathit{n}}=\frac{180}{12}=15$