Selecting books with multiple copies and posting letters to wrong envelopes

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xt
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I got this two problems, I can't figure them out...

A bookshelf contains m different books and n copies of each. How many different selections can be made from them?

and

In how many different ways can four letters be posted in four envelopes so that no one receives the correct letter?
 

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xt said:
I got this two problems, I can't figure them out...

A bookshelf contains m different books and n copies of each. How many different selections can be made from them?

and

In how many different ways can four letters be posted in four envelopes so that no one receives the correct letter?
for the first question maybe it is:
m*(m+n)

but i only checked it for two specific cases.


edit:
after checking again i think the answer should be:
if m>n than S=m*(n+m)
if m<n than S=n*(n+m)
 
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If n=1, then the answer ought to be 2^m, so which examples did you check?
 
1. For each book, you can take 0 of that book, or 1 of that book, ..., or n of that book. That is (n+1) possibilities. Since we have m books, the formula is (n+1)(n+1)(n+1)...(n+1) m times, or (n+1)^m.

One of the possibilities is that we select 0 of every book, so if "none of the books" is not a valid selection, then it is (n+1)^m - 1.
 
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matt grime said:
If n=1, then the answer ought to be 2^m, so which examples did you check?
i didnt try it for n=1.
the cases are:
m=2 n=3
and m=2 n=4
perhaps the flaw is i should have put in the second case m different than 2.
bougar :frown:
 
matt grime said:
If n=1, then the answer ought to be 2^m, so which examples did you check?
i tried with n=1 and m=2 and here are the combinations i got:
(1,1)
(1,0)
(0,1)
if you put it in your formula it equals to 2^2=4 which is one combination more than in reality.

(unless you include (0,0) a selection which i haven't included in my two examples).
 
The empty selection is always a selection.

pig's answer is correct assuming that the wording implies that picking anyone of the n copies is the same event as picking any of the other of the n copies.


The second question follows from the inc-exc principle.

Let V_n be the event that the nth person gets the right letter, U_n the even they don't

Then doing the event you want is the intersection of the V_i which is the complement of the union of the U_i, which can be found using inclusion exclusion
 
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matt grime said:
The empty selection is always a selection.
whats the logics behind this? why do you assume its a selection first then after the solution is done minus it off and say its not a valid selection?

matt grime said:
inc-exc principle.
what is that? where can i learn this stuff from? elementary probability?