Given that there are $8$ items $m, n, p, q, r, s, t,$ and $u$ will be distributed in box X and box Y, answer the following questions:
a. How many ways can you distribute the given items if either Box $X$ or Box $Y$ can have no item?
This is how I answered this problem.
Consider the equation $X + Y = 8$ where $X$ and $Y$ are non-negative integers from $0$ to $8$. We're looking for the number of solutions this equation has. Thus, we can use the start and bars method.
In this case, the total number of ways is $_9C_1$.
But accoding to my partner, the answer must $2^8$.
b. How many ways you can distribute the given items such that there must be at least $1$ item in both Box $X$ and Box $Y$?
In this case, I expressed $X$ as $X' + 1$ and $Y$ as $Y'+ 1$
Hence, $X'+1 + Y'+1 = 8$.
$X'+Y' = 6$
This follows that the number of ways to distribute is $_7C_1$.
But the answer of my partner is $2^8-2$ ways.
I am a bit confused on our answers.
Any comments or suggestions will be much appreciated. Thank you in advance.
Best Answer
the number of solutions to $X+Y=8$ does not take into account that they are distinct items. For example, having 3 items in X and 5 items in Y does not count one configuration because $X = \{m,n,p\}, Y = \{q,r,s,t,u\}$ is a different configuration you need to count separately from $X = \{q,n,p\}, Y = \{m,r,s,t,u\}$ (switched $q$ and $m$)