KVPY SA Exam 2018 Question Paper

Digits are 1, 2, 3, 4, 5

Even‌digits=2,4	;	number‌of‌Even‌digits=2
Total‌marks=2375	;	number‌of‌Odd‌digits=3

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 n=

3

O

×

2

E

×

2

O

×

1

E

×

1

O

=12
Case I For m ; Repetition is allowed
⇒ Two possipilities
(a) OEOEO
Number of such arrangements =

3

O

×

2

E

×

3

O

×

2

E

×

3

O

=108
(b) EOEOE
Number of such arrangements =

2

E

×

3

O

×

2

E

×

3

O

×

2

E

=72
So m=108+72=180

m

n

=

180

12

=15