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anonymous
 4 years ago
How many ways are there in putting 5 different books into 2 different bags so that each bag contains at least 1 book?
anonymous
 4 years ago
How many ways are there in putting 5 different books into 2 different bags so that each bag contains at least 1 book?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First put 1 book in each bag.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This 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 \[^5P_2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But am sure just plugging those in won't give the correct answer... See what you can do..

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.0Consider 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1336274424353:dw I think

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.01 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 :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0but that middle has a few duplicates that would need to be weeded out

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0ab 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.010+10+5+5 = 30 different ways ... maybe ;)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.05c1 + 5c2 + 5c2 + 5c1 = 30

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.05c1 + 5c2 + 5c3 + 5c4 = 30 might conform to the structure better tho
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