Balls and urns when urns are indistinguishable

Question

In how many ways can $100$ identical books be distributed into $10$ indistinguishable bags so that no two bags contain the same no. of books and no bag is empty?

My approach:
Though this looks like a simple balls and urns problem, it is not so. I would have easily been able to do the problem if it had not used the word "indistinguishable".
We have the equation $$\sum_{i=1}^{10} x_i=100$$ where $x_i\ge1$
So we can tackle this by replacing $x_i$ with $x_i^{'}+1$, to get $$\sum_{i=1}^{10} x_i^{'}=90$$ and then use $\binom{n+r-1}{r}$
But what to do with indistinguishable?
Edit:
So as "some SE user" told me to divide it by $10!$, I think the ans is $\frac{\binom{10+90-1}{90}}{10!}$

Indistinguishable means that “bag A containing x balls is the same as bag B containing x balls”. Consequently there will be less combinations. How many? Well: all the bags are supposed to have a different number of balls, consequently you have to factor out all permutations, i.e. divide by $10!$ — b00n heT, Dec 18 '17 at 06:30
$\binom{10+90-1}{90}$ is indeed the number of ways of putting $100$ indistinguishable items into $10$ distinguishable bags. One such example is $1+2+3+4+5+6+7+8+9+55$ and there are $10!$ permutations of that pattern; but another is $10+10+10+10+10+10+10+10+10+10$ and there is only one permutation of this pattern, so you need to exclude this and others before dividing by $10!$ — Henry, Dec 18 '17 at 07:50

Robert Z · Accepted Answer · 2017-12-18T07:55:49.863

2

Since the 10 bags are indistinguishable, the 100 books are identical, we may assume that the bags are arranged with respect to the their number of balls. Let $0< x_1<x_2<\cdots <x_{10}$ be the number of books in each bag. Then $$\sum_{i=1}^{10}x_i=100$$ that is, by letting $x_i=y_i+i-1$, $1\leq y_1\leq y_2\leq \cdots \leq y_{10}$ $$\sum_{i=1}^{10}y_i=100-\sum_{i=1}^{10}(i-1)=55.$$ So you should count the number of integer partitions of 55 in exactly 10 terms, that is $P(55,10)$ which satisfies the recurrence $$P(n,k) = P(n-1,k-1) + P(n-k,k)$$ see Integer partition of n into k parts recurrence

P.S. Your problem is asking for the number of partitions of $n=100$ into $k=10$ distinct parts. The generating function of such numbers $a(n,k)$ is $$f_k(x)=\sum_{n\ge 0}a(n,k)x^n=\frac{x^{k+\binom{k}2}}{(1-x)(1-x^2)\dots(1-x^k)}.$$ According to WA, $a(100,10)=33401$.

edited Dec 18 '17 at 07:55

answered Dec 18 '17 at 06:46

Robert Z

147,345

Can you please tell why my method is not working? – ami_ba Dec 18 '17 at 07:07
@amitayas Try to solve the problem with smaller numbers, say $10$ books in $4$ bags. What is your result in this simpler case? – Robert Z Dec 18 '17 at 07:20
No....sorry...i do not have – ami_ba Dec 18 '17 at 07:26
is b00nhrT correct? – ami_ba Dec 18 '17 at 07:27
If b00nhrT is correct, then the ans to the 10-4 problem you gave should be $$\frac{\binom{6+4-1}{6}}{4!}$$ – ami_ba Dec 18 '17 at 07:29
$x_1+x_2+x_3+x_4=10$.....$x_1^{'}+x_2^{'}+x_3^{'}+x_4^{'}=6$....if not indistinguishable ans $\binom{6+4-1}{6}$....if indistinguishable ans $\frac{\binom{6+4-1}{6}}{4!}$ – ami_ba Dec 18 '17 at 07:33
No, he is not correct. There is just ONE case: $1+2+3+4=10$. – Robert Z Dec 18 '17 at 07:36
I learnt the $X_i-X_i^{'}$ technique from https://math.stackexchange.com/questions/2559845/find-all-integer-solutions-to-the-equation-based-on-the-following-conditions – ami_ba Dec 18 '17 at 07:36
I am unable to understand the necessity of $x_i=y_i+i-1$ – ami_ba Dec 18 '17 at 07:38
Please help me out! – ami_ba Dec 18 '17 at 07:39
$x_i=y_i+i-1$ is needed because each bag contains a different number of books. Hence, in order, the first bag has at least $1$ book, the second at least $2$ books, and so on. – Robert Z Dec 18 '17 at 07:41
oh! got it now....thank you very much – ami_ba Dec 18 '17 at 08:21

score 1 · Answer 2 · answered Dec 18 '17 at 06:51

In effect, you are counting the sequences of integers $(x_1,\ldots,x_{10})$ with $1\le x_1<x_2<\cdots<x_{10}$ and $x_1+\cdots+x_{10}=100$. If we set $y_k=x_k-k$, this is the same as the number of sequences of nonnnegative integers $(y_1,\ldots,y_{10})$ with $y_1\le y_2\le\cdots\le y_{10}$ and $y_1+\cdots+y_{10}=45$. This is the number of partitions of $45$ into $\le10$ parts, or the number of partitions of $45$ into parts of size $\le10$. It is the coefficient of $q^{45}$ in $\prod_{k=1}^{10}(1-q^k)^{-1}$.

Balls and urns when urns are indistinguishable

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