Hi,palvarez wrote:you are given XXXXX; in how many ways, can you insert two |'s in it? 7C2.
XX|XX|X
X|XXX|X
The problem with your solution is that the events are dependent, not independent.vivekjaiswal wrote:Hi,palvarez wrote:you are given XXXXX; in how many ways, can you insert two |'s in it? 7C2.
XX|XX|X
X|XXX|X
May be I am really dumb or something
I am still not able to connect the dots out here...how does this explain the distribution of donuts in my question?
I had thought it would be like, one of the guy could get anything from 0 to 5 donuts. thus the first one can get a donut in 6 ways then the next in 5 and the next in 4 and thus a total of 120. Now i know this is wrong, but I would really appreciate if you could come up with a better explanation please
Cheers,
Vivek
Could someone just clarify what the difference between this problem at the one that Stuart Kovinsky answer to on the laste page of this thread: ?palvarez wrote:Treat X as donuts, the bar | as splitting.
XX|X|XX, 2 donuts for the first person; 1 for the next; 2 for the last
Now the problem is: in how many ways can you shuffle 2 '|' in between 5 donuts.
Why 2 |s? For three people in the above example, you need 2 |'s
For n people, you need n -1.
x_1 + x_2 + ... +x_k =n
where 0 <= x_i < =n
Number of solutions: (n+k-1)C(k-1) = (n+k-1)Cn. Here you see k-1 +'s.
Here is how you solve it: If x, y, and z are positive integers, and x + y + z = 9. How many combinations of x, y, and z are possible?tgf wrote:Could someone just clarify what the difference between this problem at the one that Stuart Kovinsky answer to on the laste page of this thread: ?palvarez wrote:Treat X as donuts, the bar | as splitting.
XX|X|XX, 2 donuts for the first person; 1 for the next; 2 for the last
Now the problem is: in how many ways can you shuffle 2 '|' in between 5 donuts.
Why 2 |s? For three people in the above example, you need 2 |'s
For n people, you need n -1.
x_1 + x_2 + ... +x_k =n
where 0 <= x_i < =n
Number of solutions: (n+k-1)C(k-1) = (n+k-1)Cn. Here you see k-1 +'s.
https://www.beatthegmat.com/combination-t41362.html
How is the "zeros" taken care of here? If ||XXXXX leaves all the donuts for the last person, then how is the two slots (||) in the same place counted in the combinations??
Thanks
That solution is possibly the quickest way to solve this particular problem:tgf wrote:Could someone just clarify what the difference between this problem at the one that Stuart Kovinsky answer to on the laste page of this thread: ?palvarez wrote:Treat X as donuts, the bar | as splitting.
XX|X|XX, 2 donuts for the first person; 1 for the next; 2 for the last
Now the problem is: in how many ways can you shuffle 2 '|' in between 5 donuts.
Why 2 |s? For three people in the above example, you need 2 |'s
For n people, you need n -1.
x_1 + x_2 + ... +x_k =n
where 0 <= x_i < =n
Number of solutions: (n+k-1)C(k-1) = (n+k-1)Cn. Here you see k-1 +'s.
https://www.beatthegmat.com/combination-t41362.html
How is the "zeros" taken care of here? If ||XXXXX leaves all the donuts for the last person, then how is the two slots (||) in the same place counted in the combinations??
Thanks
Right, because we don't care who gets which donuts, just who gets how many donuts.tgf wrote:Just one last question:
Whether the 5 donuts are identical or different is completely irrelevant when is it comes to the answer?
Stuart Kovinsky wrote:Right, because we don't care who gets which donuts, just who gets how many donuts.tgf wrote:Just one last question:
Whether the 5 donuts are identical or different is completely irrelevant when is it comes to the answer?
