Find the number of permutations of the multi-set {m.1,n.2}, where m,n $\in N $, which must contain m 1's.
I thought the permutation is $\frac{(m+n)!}{m!n!}$ since multi-set is basically a collection of objects with repetitions.
But the answer given is:
$\binom{m+n+1}{n}$
On
Another way to solve this is to first line up the m ones: $\;\;1\;\;1\;\;1\;\cdots\;1$.
This leaves $m+1$ gaps: $\;\;\_\;1\;\;\_\;1\;\;\_\;1\;\_\cdots\_1\;\_\;1\;\_$,
so if we let $x_i$ be the number of 2's we put in gap $i$ for $1\le i\le m+1$,
we have the inequality $x_1+\cdots+x_{m+1}\le n$ $\;\;$where $x_i\ge0$ for each $i$.
If we let $x_{m+2}=n-(x_1+\cdots+x_{m+1})$,
this gives the equation $x_1+\cdots+x_{m+1}+x_{m+2}=n$, and
this equation has $\dbinom{m+1+n}{n}$ solutions in nonnegative integers.
Since the question specified "must have $m$ 1's", it was probably meant that upto and including $n$ 2's can be there. So you need to add together the permutations for a fixed $m$ 1's and $i$ 2's for all $0 \leq i \leq n$. For a certain $n$, you stated correctly that the number of permutations are $$\binom{m+n}{n}$$ So the total number of permutations is $$\sum_{j=0}^{n}\binom{m+j}{j}=\binom{m+n+1}{n}$$ The last equation is an identity for binomials. Ask if you're unsure about where it came from :)
Edit: Here's the proof for the identity at the end. We will use induction for the variable $n$:
Base case: when $n$ equals $0$, both sides are clearly equal.
Induction: Assuming the expression holds for $n$, we have $$\sum_{j=0}^{n+1}\binom{m+j}{j}=\underbrace{\sum_{j=0}^{n}\binom{m+j}{j}}_{\binom{m+n+1}{n}}+\binom{m+n+1}{n+1} \\=\binom{m+n+1}{n}+\binom{m+n+1}{n+1}=\binom{m+n+2}{n+1}$$ which means it holds for $n+1$ which implies it holds for all $n\geq0$. In the ending we made use of the property $$\binom{n}{k-1} + \binom{n}{k}=\binom{n+1}{k}$$ which can be proven quite easily. See this.