Lottery (replacement, no order)

vashts85

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I am reading the discussion in my probability book (graduate level Mathematical Statistics I course).

The framework is a lottery with 44 numbers. Here one needs to pick 6 numbers to win.

I understand that WITHOUT replacement and ordered the chance to win is 44!/38!

I get that WITH replacement and order, the chance to win is 44^6

I also get that, unordered and with replacement, the chance to win is 44!/(6!38!) because one is eliminating repeated groupings.

What I don't understand is UNORDERED WITH REPLACEMENT. The book provides the following description, but the concept of bins does not make sense to me. I really don't understand why they go from 44 slots to 49, and then divide by 43!. I've included a picture of the book section below. I was, like the book says, guessing that the odds of winning would be 44^6/6!

Could someone help explain this to me?

32428d1421262100-lottery-replacement-no-order-untitled.png
 
I am reading the discussion in my probability book (graduate level Mathematical Statistics I course).
The framework is a lottery with 44 numbers. Here one needs to pick 6 numbers to win.
I understand that WITHOUT replacement and ordered the chance to win is 44!/38!
I get that WITH replacement and order, the chance to win is 44^6
I also get that, unordered and with replacement, the chance to win is 44!/(6!38!) because one is eliminating repeated groupings.
What I don't understand is UNORDERED WITH REPLACEMENT. The book provides the following description, but the concept of bins does not make sense to me. I really don't understand why they go from 44 slots to 49, and then divide by 43!. I've included a picture of the book section below. I was, like the book says, guessing that the odds of winning would be 44^6/6! Could someone help explain this to me?

32428d1421262100-lottery-replacement-no-order-untitled.png
Do not post the same question more that once.

You have posted this question once before. You have received replies which you clearly are not happy with. So post further questions there.
 
I am reading the discussion in my probability book (graduate level Mathematical Statistics I course).

The framework is a lottery with 44 numbers. Here one needs to pick 6 numbers to win.

I understand that WITHOUT replacement and ordered the chance to win is 44!/38!

I get that WITH replacement and order, the chance to win is 44^6

I also get that, unordered and with replacement, the chance to win is 44!/(6!38!) because one is eliminating repeated groupings.

What I don't understand is UNORDERED WITH REPLACEMENT. The book provides the following description, but the concept of bins does not make sense to me. I really don't understand why they go from 44 slots to 49, and then divide by 43!. I've included a picture of the book section below. I was, like the book says, guessing that the odds of winning would be 44^6/6!

Could someone help explain this to me?

32428d1421262100-lottery-replacement-no-order-untitled.png
Not finding the duplicate post anywhere, I will attempt to answer the question here.

IMO, a good understanding of combinations (order isn't important) without replacement and permutations (order is important) with and without replacement is needed in order to understand combinations with replacement. Every once in a while, I seem to have enough understanding to answer one of those kind of questions but most of the time, I just point people to
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
for a good, IMO, introduction to the subject.
 
Not finding the duplicate post anywhere, I will attempt to answer the question here.

IMO, a good understanding of combinations (order isn't important) without replacement and permutations (order is important) with and without replacement is needed in order to understand combinations with replacement. Every once in a while, I seem to have enough understanding to answer one of those kind of questions but most of the time, I just point people to
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
for a good, IMO, introduction to the subject.

Thank you, I will take a read at this. I had posted this question to another forum; perhaps you saw it there and thought I had double-posted here.

Thanks for trying to help !! Will reply with feedback on the link.
 
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