Here's a Permutation/Combination that stymied even my Math teacher;?

Question

In how many ways can 5 different books be distributed among 10 people, if each person canget any number of books?

I can't really solve it, but I think it shouldn't be too hard...

Answer 1

it's the coefficient of x^5 in (1 + x + x^2 + ...+x^5)^10
= coeff of x^5 in (1 - x)^(-5)
= 10*11*12*13*14/(1*2*3*4*5)
=2002 ways is the correct answer

Answer 2

These are the possible combinations
(i) 1 person gets all the books - 10C1 = 10 ways.
(ii) 2 people get the books -
P1:1 +P2: 4 - 10 * 9 = 90 ways
P1:2 +P2: 3 - 10 * 9 = 90 ways
Since it is combinations, we need not consider the case of 3+2 and 4+1, this has been taken care of by the first case.
(iii) 3 people get the books.
1 + 1 + 3 : 10*9*8 = 720 ways
1 + 2 + 2 : 10*9*8 = 720 ways
(iv) 4 get the books
1 + 1 + 1 + 2 : 10*9*8*7 = 5040 ways
(v) 5 books go to a single person - 10 ways
Thus the total # ways is 6680 ways

Answer 3

Hi, think of it this way. You have 5 items that you can put any way into 10 boxes. Let * be a book and | be a divider between two boxes.
So, *****| | | | | | | | |, would represent all 5 books in the first box,
[note: you need just 9 | to make 10 sections]
and | | | | | | | | | *****, would be all 5 books in the last box.
So, in our little diagram we have 14 symbols. Rearranging them gives us 14!, but rearranging the five books doesn't really matter, and rearranging the dividers among themselves doesn't matter so we need to divide out 5! and 9!
So, the answer is 14!/(5!9!) = 2002.

In general then, n books to p people would be
(n+p-1 choose n)

Answer 4

The answer, in my opinion should be 2002.
Since any student can have any number of books, we will try all combinations.
1 STUDENT (05):
Single student has all 5 books. i.e say student A has 5 books. Since any of 10 student could be A we have (10 choose 1)= 10 ways for 05 combination. We are not required to think of 50 combination since at a time ONLY SINGLE student is considered.
2 STUDENT combination(14, 41, 23, 32)
Let students A and B be selected such that distribution of books is such done as AB=14(ie 1 book for A and 4 for B); AB=41; AB=23; AB=32. Since students AB could be any of 10 students there are (10 choose 2)=45 ways of selecting students and FOUR ways for EACH in which books could be distributed among TWO students. Hence total ways=45x4=180 ways.
3 STUDENT combination(ABC= any of113, 311, 131, 122, 212, 221):
Ways in which books could be distributed among three students at the same time = 6 ways.
Ways in which 3 could be selected out of 10 is (10 choose 3)=120 ways(Note no need to take permutations as those are taken care of right in the beginning when distribution among 3 was considered.) Total no. of ways=6x120=720 ways
4 STUDENT combination(ABCD=1112, 1121, 1211 and 2111):
Among 4 students books could be distributed as above combinations(total four for EACH ABCD combination)
Number of ways in which 4 among 10 are selected are (10 choose 4)=210 ways.
Total ways=4x210=840 ways.
5 STUDENT combination(ABCDE=11111)
No other combination other than above is possible. Look carefully here you will understand why its not required to calculate permutations.
Total no of ways=1x(10 choose 5)=252 ways.
(Obviously, since number of students exceed number of books, atleast 5 will be left without books.)
Grand total ways=10+180+720+840+252=2002 ways
Good question!

Answer 5

It's simple 2 people for each book of the five.... 5 books = 10 people = 2 people per book.

Answer 6

The first book gets a choice of 10 people - so does the second, third, fourth and fith book so its 10 X 10 X 10 X 10 X 10 = 100,000 different ways.

Answer 7

10^5, for the reason previously stated. Each book can go to one of 10 people; and the assignments of books to people are independent of each other.

Answer 8

Commentary of previous answers: SH_M_C makes sense if the books are indistinguishable. But if you can tell them apart (and tell the people apart), then I think Jinoturi is correct -- nice easy way of looking at the problem.

Answer 9

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Answer 10

a permutation is an ordered list without repetition, perhaps missing some elements. I think that fits your description.

answer = 10P5 = 30,240