I'm just trying to understand how to use PIE one word problems. I've got this silly situation in front of me:
A woman is having a dinner party with 9 of her female friends, and she's every so nicely written each lady's name on a piece of paper and placed it at a seat at her kitchen table. Unfortunately, her dough-head of a husband left the back door open! It turned out to be a very blustery day, and the oh so blustery wind came in and blew all the seat assignments away! And now, the husbands flustered and trying to recover the seat assignments... How many ways are there for him to arrange the pieces of paper so that 1) exactly 4 of the 10 ladies are sitting in the spots that the wife intended? and 2) at least 4 of the 10 ladies are sitting in the spots that the wife intended?
So here's what I've come up with:
We've got 10 women, $w_1, w_2, .. , w_{10}$. Let $c_i$ be the condition that $w_i$ (where $1 \leq i \leq 10$) is seated in the spot that the wife originally intended.
We let $S_j$ be the set of arrangements that has $j$ (where $0 \leq j \leq 10$) women being seated in the spot originally intended for them.
When determining the value of $S_j$, we first need to choose $j$ seats from $10$, then arrange the remaining $10-j$ individuals around the table.. So we have:
$S_j = \binom{10}{j}*(10-j)!$
Then, the answer to 1) will be
$S_4 - S_5 + S_6 - S_7 + S_8 - S_9 + S_{10}$
$= \binom{10}{4}(6!) - \binom{10}{5}(5!) + \binom{10}{6}(4!) - \binom{10}{7}(3!) + \binom{10}{8}(2!) - \binom{10}{9}(1!) + \binom{10}{10}(0!)$
$= 125361$
The answer to 2) is simply $S_4$ and does not require the use of PIE.
Can anyone comment on whether I've gone about this in the right way?
Cheers!