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First put 1 book in each bag.
This is a long process without formula... however, if you want to be certain, try formulas with Combinations and Permutations, and list out all the ways if you have time.. Head start on the listing: 5 different books = a,b,c,d,e Bags= B1, B2 B1~~B2 a ~~ b a ~~ c a ~~ d a ~~ e ab~~c ab~~d ab~~e and so forth.. Not sure with the formulas, but you could try something like \[^5P_2\]
But am sure just plugging those in won't give the correct answer... See what you can do..
Consider bag A. The number of combinations of books taken one at a time = 5!/4! =5 The number of combinations of books taken two at a time = 5!/(2*3!) = 10 The number of combinations of books taken three at a time = 5!/(3!*2!) = 10 The number of combinations of books taken four at a time = 5!/4! = 5 The total number of allowed combinations of books in bag A = 5 + 10 + 10 + 5 = 30 The total number of allowed combinations of books in bag B will also equal 30. Therefore the total number of ways of putting the five different books into two different bags so that each bag contains at least one book = 30 + 30 = 60 ways
|dw:1336274424353:dw| I think
1 4; 5ways 2 3 =*(1, 3) 4 ways 5 times 3 2 ; same as above 4 1; same as above 2*5 + 2*(4*5) = 10 + 40 = 50 maybe :)
but that middle has a few duplicates that would need to be weeded out
ab cde ac bde ad cbe ae dcb bc dea bd cea be dca ba cde ** duplicate cd eab ce dab ca edb ** duplicate cb dea ** duplicate de abc da ** db ** dc ** ea ** eb ** ec ** ed ** 10 ways ... to stack 2,3
10+10+5+5 = 30 different ways ... maybe ;)
5c1 + 5c2 + 5c2 + 5c1 = 30
5c1 + 5c2 + 5c3 + 5c4 = 30 might conform to the structure better tho