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Can anyone help me with this combination question?
There were 10 different Mathematics books in the library.
(i) In how many different ways can the books be distributed equally among 5 friends?
(ii) How many ways can the 10 books be put into 5 piles of 2 each on the table?
(iii) If one student were to get 1 book, another to get 2 books, the third to get 3 books, the forth 4 books and the fifth nothing, how many ways can you distribute the 10 books?
The answer is....
(i)10C2 x 8C2 x 6C2 x 4C2 x 2C2 = 113400
(ii) 113400 / 5! = 945
(iii)10C1 x 9C2 x 7C3 x 4C4 x 5! = 1512000
But how come question one and two are not same? Why do i need to divided by 5!?
and I am confusing question 3 also.
2 Answers
- MadhukarLv 71 decade agoFavorite Answer
The reason is that 5 friends are distinguishable whereas 5 files are not. So mutual arrangement of 5 piles is meaningless. Hence division by 5!
Explaining in another way, there is the first friend, second friend, etc. But there is nt the first file, second files, etc. It means that all files are equivalent whereas all friends are not.
To make things more clear, let us compress the problem to two friends M and N and four books A, B, C, and D. This can be divided amonst two frinds as under
M - AB and N - CD
M - AC and N - BD
M - AD and N - BC
M - BC and N - AD
M - BD and N - AC
M - CD and N - AB
Total 4C2 x 2C2 = 6
Now, 4 books can be made into 3 piles only because AB and CD is same as CD and AB, AD and BC is the same as BC and AD and BD and AC is the same as AC and BD.
Total 4C2 x 2C2 / 2! = 3.
- Anonymous1 decade ago
The reason questions one and two are not the same is that in question one the five friends are distinct, but in question two the five piles are indistinguishable. In other words, once you have distributed the books into the five piles you can move the piles anywhere on the table, but that doesn't affect the distribution. However, if Anne and Bob switch books with each other, that yields a different distribution.
The same reasoning also explains why you multiply by 5! in question three. There are 5! ways to decide who gets the one book, who gets two, etc.
