probability with combination

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achal

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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 have found that the number of such permutations will be (m1+1)m2(m3-1)......(mk-k+2), but what will be the number of combinations ? Clearly it can not be just = No. of permutations / k!

N.B. - Please mention the formula/rules you've used so that I can learn them.
 
achal said:
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 have found that the number of such permutations will be (m1+1)m2(m3-1)......(mk-k+2), but what will be the number of combinations ? Clearly it can not be just = No. of permutations / k!
Click to expand...
I don't think that you really understand this question.
Take an example: \(\displaystyle 2<4<5<7,~k=4\).
Then \(\displaystyle M_1=\{0,1,2\},~M_2=\{0,1,2,3,4\}\)\(\displaystyle ,~M_3=\{0,1,2,3,4,5\},~\&~M_4=\{0,1,2,3,4,5,6,7\}\).
There are three ways to choose one from \(\displaystyle M_1\).
Then there are three ways to choose one from \(\displaystyle M_2\).
Then there are four ways to choose one from \(\displaystyle M_3\).
Then there are five ways to choose one from \(\displaystyle M_4\).
That is 180 choices. Now it must be done in that order.
Otherwise we may not be able to complete the task.
For example: doing this way \(\displaystyle 0\in M_4,~2\in M_3,~1\in M_2\) does not leave any choice from \(\displaystyle M_1\)
So permutations do not even come into this at all.

I have no idea right now how to count the disjoint pairs of choices.
 
pka said:
I don't think that you really understand this question.
Take an example: \(\displaystyle 2<4<5<7,~k=4\).
Then \(\displaystyle M_1=\{0,1,2\},~M_2=\{0,1,2,3,4\}\)\(\displaystyle ,~M_3=\{0,1,2,3,4,5\},~\&~M_4=\{0,1,2,3,4,5,6,7\}\).
There are three ways to choose one from \(\displaystyle M_1\).
Then there are three ways to choose one from \(\displaystyle M_2\).
Then there are four ways to choose one from \(\displaystyle M_3\).
Then there are five ways to choose one from \(\displaystyle M_4\).
That is 180 choices. Now it must be done in that order.
Otherwise we may not be able to complete the task.
For example: doing this way \(\displaystyle 0\in M_4,~2\in M_3,~1\in M_2\) does not leave any choice from \(\displaystyle M_1\)
So permutations do not even come into this at all.

I have no idea right now how to count the disjoint pairs of choices.
Click to expand...

No, there can be other way: say m1 = 5, m2 = 10 and m3 = 17
then a combination, say {1,3,4} can be made by either taking 1 from M1, 3 from M2 and 4 from M3
OR 2 from M1, 4 from M2 and 1 from M3 etc in 3! ways. so there comes the question of permutations and not all combination has k! permutations.
 
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