Kaederfds
  • Kaederfds
How many ways are there in putting 5 different books into 2 different bags so that each bag contains at least 1 book?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
First put 1 book in each bag.
anonymous
  • anonymous
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 \[^5P_2\]
anonymous
  • anonymous
or \[^5C_2\]

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anonymous
  • anonymous
But am sure just plugging those in won't give the correct answer... See what you can do..
kropot72
  • kropot72
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
Kaederfds
  • Kaederfds
|dw:1336274424353:dw| I think
amistre64
  • amistre64
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 :)
amistre64
  • amistre64
but that middle has a few duplicates that would need to be weeded out
amistre64
  • amistre64
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
amistre64
  • amistre64
10+10+5+5 = 30 different ways ... maybe ;)
amistre64
  • amistre64
5c1 + 5c2 + 5c2 + 5c1 = 30
amistre64
  • amistre64
5c1 + 5c2 + 5c3 + 5c4 = 30 might conform to the structure better tho

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