Question

Peter Parker gets 5 coupons from a restaurant. All the coupons have to be used within upcoming 120 days and he CANNOT use two or or more coupons in one day. He plans to visit the restaurant 5 times in 120 days. But after each visit, he will NOT visit the restaurant in next two days. For example, if he visits the restaurant on 1st of September, he will NOT visit it on 2nd and 3rd of September. How many ways can he choose the 5 days?

Answer 1

120C5 is the total number of time he can go

so out of if he go continiouslyevery day then it is 116C1 because it can happen only till 116 i.e (116,117,118,119,120)

if two goes together and the rest three are at random i,e116C2

similarly

116C3

116C4

now given gaps of two we have take care of the other version i.e of 60

then 56C1

56C2

56C3

56C4

hence the answer is 120C5-(116C1+116C2+116C3+116C4)-(56C1+56C2+56C3+56C4)

> choose(120,5)-sum(choose(56,c(1,2,3,4)))-sum(choose(116,c(1,2,3,4)))
[1] 182760927