# probability with combination

• Dec 16th 2011, 02:57 AM
achal
probability with combination
There are k sets of numbers : {0,1,2,….,m1}, {0,1,2,……..,m2}, …………,{0,1,2,………,mk}
Such that m1<m2<………<mk.
1. How many combinations of k elements can be made taken 1 element from each set such that each set has all distinct elements (no two elements are equal) ?
2. What is the probability that any two sets will have at least one element common ?

I think the answer given by Plato is the number of permutations, but I am looking for the number of combinations i.e., the sets with all elements same (although in different order) will be treated as one.

AND Plz mention the rule/formula name with explanation so that I can learn them
• Dec 16th 2011, 07:31 AM
Plato
Re: probability with combination
Quote:

Originally Posted by achal
There are k sets of numbers : {0,1,2,….,m1}, {0,1,2,……..,m2}, …………,{0,1,2,………,mk}
Such that m1<m2<………<mk.
1. How many combinations of k elements can be made taken 1 element from each set such that each set has all distinct elements (no two elements are equal) ?

I for one find this question hard to follow.
Here is the way I read it.
There is a increasing sequence of integers: $0.
There are sets $A_n=\{1,2,\cdots,m_n\}$.
We pick a number from $A_1$, then we pick a different number from $A_2$, etc until we pick yet a different number from $A_k$. Thus we have $k$ different selections.

The question seems to be, "How many different selections are possible?"

Note the the number of integers in each set is $\|A_n\|=m_n+1.$
If this is the correct setup, the answer is:
$\prod\limits_{n = 1}^k {\left( {m_n + 2 - n} \right)}$