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 ?

(Please provide procedure and explanation)

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

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: $\displaystyle 0<m_1<m_2<\cdots<m_k$.

There are sets $\displaystyle A_n=\{1,2,\cdots,m_n\}$.

We pick a number from $\displaystyle A_1$, then we pick a **different number** from $\displaystyle A_2$, etc until we pick yet a **different number** from $\displaystyle A_k$. Thus we have $\displaystyle k$ different selections.

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

Note the the number of integers in each set is $\displaystyle \|A_n\|=m_n+1.$

If this is the correct setup, the answer is:

$\displaystyle \prod\limits_{n = 1}^k {\left( {m_n + 2 - n} \right)} $