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Kaederfds

  • 2 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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  1. estudier
    • 2 years ago
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    First put 1 book in each bag.

  2. order
    • 2 years ago
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    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 \[^5P_2\]

  3. order
    • 2 years ago
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    or \[^5C_2\]

  4. order
    • 2 years ago
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    But am sure just plugging those in won't give the correct answer... See what you can do..

  5. kropot72
    • 2 years ago
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    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

  6. Kaederfds
    • 2 years ago
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    |dw:1336274424353:dw| I think

  7. amistre64
    • 2 years ago
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    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 :)

  8. amistre64
    • 2 years ago
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    but that middle has a few duplicates that would need to be weeded out

  9. amistre64
    • 2 years ago
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    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. amistre64
    • 2 years ago
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    10+10+5+5 = 30 different ways ... maybe ;)

  11. amistre64
    • 2 years ago
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    5c1 + 5c2 + 5c2 + 5c1 = 30

  12. amistre64
    • 2 years ago
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    5c1 + 5c2 + 5c3 + 5c4 = 30 might conform to the structure better tho

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