Question

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

Answer 1

First put 1 book in each bag.

Answer 2

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 \[^5P_2\]

Answer 3

or \[^5C_2\]

Answer 4

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

Answer 5

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

Answer 6

I think

Answer 7

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 :)

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

5c1 + 5c2 + 5c3 + 5c4 = 30

might conform to the structure better tho