JOMO 5, Short 9
One day, 7 friends decided to find a rare Mathematics book (Imagine the ancient Fermat’s notebook). To do so, they sourced out 5 oldest bookshops in the world. However, due to time constraint (they have to get the book before someone else does), they decided to split themselves up such that there is at least 1 person who visits each bookshop. Find the number of ways that they can arrange themselves divided by 100.
Details and Assumptions :-
The 7 friends are considered distinct people. Same applies for the 5 bookshops
The answer is 168.
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1 solution
take 5 book centers say....a,b,c,d,e...........and 7 persons .......
now....there are two ways of various people visiting the shops with each shop is visited by atleast one person.....
3 1 1 1 1; 2 2 1 1 1
in the first case......if we describe...any 3 of 7 visit the first then 1 each to all other four shops....mathematically....7c3 4 3 2 1.....now considering shops combination answer is multiplied by 5!/4! (as four solutions are similar 1 1 1 1).that makes no of possible ways.
7c3 4 3 2 1*(5!/4!)=4200
similarly in the second case...no of ways
7c2 5c2 3 2 1 (5!/(3! 2!))=12600
total ways=4200+12600=16800 and that devided by 100 is 168...