From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to ab, where a, b N and (a, b)=1, then a+b is equal to .
Answer
944
Solution
Total ways =31C3= 31 30 296=4495
a<b<c are in A.P. b= a+c2, so a, c must be both odd or both even (then b is fixed).
Odd dates: 16 16C2=120; even dates: 15 15C2=105
Favourable =120+105=225
P= 2254495= 45899 (899=29 31, so (45, 899)=1)
a+b=45+899=944
Answer: 944