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## Kaederfds 3 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

First put 1 book in each bag.

2. order

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

or \[^5C_2\]

4. order

But am sure just plugging those in won't give the correct answer... See what you can do..

5. 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

6. Kaederfds

|dw:1336274424353:dw| I think

7. 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 :)

8. amistre64

but that middle has a few duplicates that would need to be weeded out

9. 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

10. amistre64

10+10+5+5 = 30 different ways ... maybe ;)

11. amistre64

5c1 + 5c2 + 5c2 + 5c1 = 30

12. amistre64

5c1 + 5c2 + 5c3 + 5c4 = 30 might conform to the structure better tho

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