Question

In how many ways 10 different books can be divided into 2 parts?

Answer 1

is that 5C10 ??

Answer 2

do u want to make 2 "equal" groups ?

Answer 3

if so, then \(\large ^{10}C_5\) is the right answer,
but in the question it doesnt speify they need to be "equal" right ?

Answer 4

nope , not 2 equal groups. The total numbers of ways we can divide the 10 differnt teachers, it can be like
2 teachers + 8 teachers group etc.

Answer 5

sorry books , not teachers

Answer 6

so, u can have below group sizes :-

group 1 : 0 books
group 2 : 10 books


group 1 : 1 books
group 2 : 9 books

group 1 : 2 books
group 2 : 7 books


....


group 1 : 10 books
group 2 : 0 books

Answer 7

right ?

Answer 8

yup and since these r differnt books then
(book 1, book2 )+(book3 . . book 10)
group is different then the
(book 2 , book 3) +(book1 , book 4 . . book 10)

Answer 9

but
(book 1, book2 )+(book3 . . book 10)
and
(book2, book1)+(book3 . . book 10)
is same , the order doesnt matters

Answer 10

i didnt get wat u wrote above :|

u should get the total ways as below :-

\(\large ^{10}C_0 + ^{10}C_1 + ^{10}C_2 ~~ ...~~ ^{10}C_{10} \)

Answer 11

uhmm see if that looks okay

Answer 12

I am saying that , if the 10 books are like
b1, b2,b3,b4,b5,b6,b7,b8,b9,b10
then if u make the 2 parts of them , then
possible ways are as u said, (2 books , 8 books) and so on
but in this case too we have other cases like
(b1,b2),(b3. . b10)
(b3,b2)(b1,b4,b5,b6. . b10)

Answer 13

wait a sec

Answer 14

sure :)

Answer 15

I am saying that , if the 10 books are like
b1, b2,b3,b4,b5,b6,b7,b8,b9,b10
then if u make the 2 parts of them , then
possible ways are as u said, (2 books , 8 books) and so on

Answer 16

I am okay till above line :)

Answer 17

yup

Answer 18

possible ways for SIZE of books in each group are :
(0, 10)
(1, 9)
(2, 8)
...
(10, 0)

Answer 19

fine till here right ?

Answer 20

Yup absolutely

Answer 21

good, next we simply find the total "ways for each SIZE"

Answer 22

total ways of having (0, 10) SIZE :-

\(\large ^{10}C_0\)

Answer 23

still fine ha ?

Answer 24

[Math Processing Error] ??
what is that ?

Answer 25

total ways of having (0, 10) SIZE :-

[Math Processing Error]

Answer 26

ahh looks latex is not working, u are not seeing what im seeing on my screen.
let me take a snapshot of wat im seeing, and attach u :)

Answer 27

yeah i cudnt see 10C0

Answer 28

but fine i got u , next ?

Answer 29

next find the "number of ways of choosing groups such that SIZEs are (2, 10)"

Answer 30

thats simply 10C2 right ?

Answer 31

yup :)

Answer 32

next find the "number of ways of choosing groups such that SIZEs are (3, 10)" :

10C3

Answer 33

keep going...

Answer 34

total number of ways of dividing 10 books into 2 parts wud be :-

10C0 + 10C1 + 10C2 + ... 10C9 + 10C10

Answer 35

umm is this the answer ?
dnt we need to again see the combinations inside the groups we created ?

Answer 36

thats the final answer.

lets consider this case : (3, 7)

Answer 37

first group : 3 books
second group : 7 books

Answer 38

yup

Answer 39

how many ways u can choose 3 books for the first group ?

Answer 40

10 C 3

right ?

Answer 41

7C3

Answer 42

opps sorry ya 10C3

Answer 43

my bad

Answer 44

okay good :)

the important question now :-
does the "order" in which u choose the 3 books matter here ?

Answer 45

the order doesnt matters but the book number does matters , as the books are different

Answer 46

exactly, and u can get 10C3 DIFFERENT book numbers of size 3

Answer 47

if we include book 1 ,book 2, book 3 in one group of (3,7)
and in another book 3, book 4 , book1 in another group of (3,7)
then it is differnt, and we need to consider that case too, will it be covered in 10C3?

Answer 48

yup !

10C3 precisely includes such cases oly

Answer 49

10C3 means, number of ways of "choosing" 3 different books from a group of 10

Answer 50

ohh nice and what about the remanining 7 books ? will it consider that too ?

Answer 51

10C3 includes below :-

book1, book2, book3
book1, book2, book4
book1, book2, book5
book10, book9, book8

Answer 52

we dont care wat happens in other group cuz :-
if u select 3 books in ur group, the remaining 7 HAVE to go into other group.

Answer 53

10C3 = 10C7

Answer 54

u must be knowing the identity :-

nCr = nC(n-r)

Answer 55

yup . I got you !! Thanks a lot :)

Answer 56

np :)

Answer 57

Also u may want to divide the final answer by 2, cuz the order doesnt matter for the two DIVISIONs as well. for example :-

(3, 7) and (7, 3) should be considered same.

Answer 58

guess id let think about it.. :)

Answer 59

ya u r correct but why divide it by 2?

Answer 60

cuz we want to treat below two SIZEs as same :-

(3, 7) = (7, 3)

Answer 61

but we have counted both earlier right ?

Answer 62

yeah 10C3 and 10C7

Answer 63

we can either divide by 2 or simply add only below :-

10C0 + 10C1 + 10C2 + .... 10C5

Answer 64

ohk so , 10C0 + 10C1 . . + 10C5 is same as
(10C0+10C1 . . . . 10C10)/2 . . Right ?

Answer 65

i thought it was the same, but let me think a bit more lol, they dont look same :o

Answer 66

but the final answer should be :

10C0 + 10C1 + .... 10C5

Answer 67

lets see if that equals below or not :
(10C0+10C1 . . . . 10C10)/2

Answer 68

ohk :)

Answer 69

I think it's same

Answer 70

(10C0+10C1 . . . . 10C10)/2

(10C0 + 10C10 + 10C1 + 10C9 + .... 10C4 + 10C6 + 10C5)/2

(2*10C0 + 2*10C1 + .... 2*10C4 + 10C5)/2

Answer 71

nope, 10C5 did not get a pair lol, so they're not a same !
thanks for asking the question lol, its a very good question :)

Answer 72

(10C0+10C1 . . . . 10C10)/2

(10C0 + 10C10 + 10C1 + 10C9 + .... 10C4 + 10C6 + 10C5)/2

(2*10C0 + 2*10C1 + .... 2*10C4 + 10C5)/2

10C0 + 10C1 + .... 10C4 + (10C5)/2

Answer 73

he he my pleasure I am stuck at this question since many days , well i think 10C5 is out of the box stuff here

Answer 74

exactly !

so the final answer is :-
10C0 + 10C1 + 10C2 + ... + 10C5

Answer 75

but if u consider the first and second groups as different, the final answer wud be :-
10C0 + 10C1 + 10C2 + .... + 10C10

Answer 76

we just need to see that both of them are different... give watever answer ur professor wants :)

Answer 77

yup , Thanks a lot for engaging so actively in this question . I wud definately have struck on this ques if u wudnt hv helped me. So Thanks a ton !!

Answer 78

np.. u wlc :)

Answer 79

Not sure I would consider something a group if it has no members. If I had 10 students and I told them to get into 2 groups and they formed a group of 10 I would not be happy ;)

Answer 80

also, recall that

\[\Large\sum_{k=0}^{n}{n\choose k}=2^n\]

Answer 81

So what do you think we should do instead ?

Answer 82

Zarkon wants us to remove 10C0 case, as 0 books in first group makes no sense

Answer 83

ohk so it should be like 10C1 + 10C2+10C3+10C4+10C5
is it ?

Answer 84

yes i think thats what Zarkon meant, it does make sense if we think a bit... but again ur professor is the boss

Answer 85

Hmm , I will ensure that :)