1

if we have 3 types of objects A,B,C . If I want to make permutation without repeat

n! = 3! = 6

but if i will take r sample so 

n!/(n-r)!

if I want to make arrangment of undistinguishable Objects like 2A 2B 2C so I may use 

n!/A!B!C!

how to make the same last test if I will take r samples

youssef
  • 111

1 Answers1

0

If I understand right, you are asking: we have $N$ items grouped in $M$ groups $n_1,n_2 \cdots n_M$, $\sum n_i=N$, the items inside each group are undistinguishable . We want to count the number of ordered selections of $R\le N$ items .

Let $r_i$ be the number of elements selected from group $i$, $0\le r_i \le n_i$, $\sum r_i=R$. Then he number of arrangements is

$$ \sum_{ 0\le r_i \le n_i\atop \sum r_i=r }\frac{r!}{r_1! r_2! \cdots r_m!}$$

Not very nice -even if $n_i=n$ is constant- I don't know if it can be simplified. If $n_i=n$ is constant ($N=nM$), a recursion (for fixed $n$) could be useful, at least for numerical computation:

$$C^n_{M,R}=C^n_{M-1,R}+{R \choose 1} C^n_{M-1,R-1}+{R \choose 2} C^n_{M-1,R-2}+\cdots =\sum_{k=0}^{ \min(R,n)}{R \choose k} C^n_{M-1,R-k} $$

leonbloy
  • 63,430
  • I don't know the sample size from each group, I will take r samples in a random way. – youssef Jan 29 '15 at 13:02
  • Of course, $r_i$ (the sample size for each selection) is not fixed, what is fixed is $n_i$ (the available items in each undistinguishable group) – leonbloy Jan 29 '15 at 13:49