Ask for my friend, I don't know this either. In how many ways can 5 letters be mailed if there are: a) 2 mailboxes available b) 4 mailboxes available Edit: I erroneously wrote that 5 mailboxes is available in b).
My intuition is that the answer to the first one is 6 as: A B 0 5 1 4 2 3 3 2 4 1 5 0 But the answer is 25. Apparently the letters are distinguishable.
@Kainui HELP!!!!!!
If the letters are distinguishable then each letter has 2 possible states : {box1, box2} So total possible states for sending 5 letters would be : 2*2*2*2*2 = 2^5
if the letters are distinguishable I get 32 ways, not 25
thats for part a
Me too. I am so confused right now.
The answer to part a is 25 and part b is 625 = 25*25.
The book could well be wrong. It happened before but I need a solid answer to say the book is wrong.
Is it possible that the mail is indistinguishable but some mail can be lost in post? LOL!
list them all if that states arguement doesn't look solid
call the letters : l1, l2, l3, l4, l5 and boxes : b1, b2
It is basically \[ \sum_{k=0}^5\binom{5}{k}=2^5 \]
thats the total number of ways of choosing 0->5 letters for first box second box is determined after that so yeah
How do you deal with the second part though? 4^5?
Yes assuming the letters are distinguishable
If they're not distinguishable then the answer is simply 6 for part a and 56 for part b
oh, it looks like the answer is doing (# of letters)^(# of boxes) so 5^2 = 25 and 5^4= 625
The book is dumb? \[ 2^5=32\\ 5^2=25\quad\text{book's answer to part a}\\ 4^5=1024\\ 5^4=625\quad\text{book's answer to part b} \]
so the book is pretty close
the author must be drunk when writing solution
*facepalm* I wouldn't call that pretty close. Yes the author must be drunk.
It is beyond words!
they used both numbers, and an exponent.
whats the complete question anyways
That is the exact wording of that question LOL!
I know the question is worded poorly and it is pretty common in this textbook.
im just wondering if there is a way to get to textbook answer by putting constraints like 1) making letters indistinguishable 2) each box must have atleast 1 letter etc..
If those constraints are true, then part a would be 4 not 25.
Join our real-time social learning platform and learn together with your friends!