Satisfiability Problem: Determining Which People To Invite 
When planning a party you want to know whom to invite. Among the people you would like to invite are three touchy friends. You know that if Jasmine attends, she will become unhappy if Samir is there, Samir will attend only if Kanti will be there, and Kanti will not attend unless Jasmine also does. Which combinations of these three friends can you invite so as not to make someone unhappy?

Okay, this problem gives three different facts that express the relationship between things occuring/being true; and here is how I translated them
(1) If Samir attends the party, then Jasmine will be unhappy.
(2) If Samir attends the party, then Kanti will be there.
(3) If Kanti attends the party, then Jasmine must also attend the party.
I correctly translated (1) and (3); however, the answer key claims that (2) should be written as 
(2) If Kanti attends the party, then Samir must attend the party.
This does not seem consistent with the answers I received here. If I read the posts correctly, $p$ only if $q$ translates to $p \implies q$
 A: Denote by $P_X$ the proposition that person $X$ will attend the party. Then from the story we conclude that
$$P_S\Rightarrow P_K\Rightarrow P_J\Rightarrow\neg P_S\ .\tag{1}$$
This chain implements in turn "$S$ will be there only if $K$ is there", "$K$ will not attend unless $J$ does", and "$J$ will not attend if $S$ does".
One of $P_S$ and $\neg P_S$ is true, so there is a first term in $(1)$ which is true. $P_S$ can definitely not be true. When $P_K$ is true, $K$ and $J$ attend; when $P_J$ is the first true term only $J$ attends, and when $\neg P_S$ is the first true term none of the three attends.
A: The problem says that if Kanti doesn't attend the party then Samir will not attend the party. So if Kanti attends the party, Samir may or may not attend the party.
So you have to invite Jasmine and Kanti only so that no one is unhappy.
A: Start with rewriting the three conditions as :
(1) if Samir will attend, then Jasmine will become unhappy ,
(2) if Kanti will attend, then Samir will attend ,
(3) Jasmine will attend or Kanti will not attend .
Let :
$U_J$ :=" Jasmine will become unhappy"
$K$ := Kanti will attend
$S$ := Samir will attend
Then the three conditions become :
(1) $S \rightarrow U_J$
(2) $K \rightarrow S$
(3) $J \lor \lnot K$ i.e. $K \rightarrow J$.
Using Tableaux method, of the eight possibilities two are unsatisfiable and four imply the unhappiness of Jasmine. So only two possibilities are left : both exclude Kanti and Samir.
You must invite Jasmine only. 
A: I'm too new to Mathematics Stack Exchange, so instead of using special symbols, I will just answer in plain text. Please feel free to edit my answer to make it conform to community's style guide.
I believe that the OP has no problem to solve the problem. It's just the problem of the interpretation of the word only if, as the OP wrote in the question, "If I read the posts correctly, p only if q translates to p -> q". And I agree with the OP.
"p only if q" means p -> q, not q -> p. The answer key seems to interpret only if as if it were just an if. -- The If and only if page in Wikipedia clarifies the distinction clearly, under the section Distinction from "if" and "only if", and it confirms that the OP is right.
To solve the problem, let J, S, K mean we can invite Jasmine, Samir, and Kanti, in that order. According to the problem,

1) You know that if Jasmine attends, she will become unhappy if Samir is there
  2) Samir will attend only if Kanti will be there, and
  3) Kanti will not attend unless Jasmine also does.

To fulfill the conditions (not to make someone unhappy), the three can be rewritten to:

1) S -> !J
  2) S -> K
  3) K -> J

Let's consider the conditions, one case at a time (0 = false, 1 = true):

J S K
  0 0 0
  0 0 1 -- fails 3
  0 1 0 -- fails 2
  0 1 1 -- fails 3
  1 0 0
  1 0 1
  1 1 0 -- fails 1, 2
  1 1 1 -- fails 1

This leaves us with only 3 possible choices:


*

*invite only Jasmine,

*invite both Jasmine and Kanti, or

*invite no one at all.

