The following problem is from Carol Ash's The Probability Tutoring Book
If $4$ people are assigned seats at random in a $7$ seat row, what is the probability that they are seated together?
My approach:
We can choose $4$ from $7$ seats: $_{7}C_4$ ways,
We can permute the people in $4!$ ways.
So, my answer was $\frac{4!}{_{7}C_4}$
However the answer in the book is $\frac{4}{_{7}C_4}$. did they miss the factorial? Or is this the correct answer?
Why is the numerator just $4$?
Is it because if I permute the numerator, I also have to permute the denominator? Can someone explain why this is logically so?
Best Answer
You have to be clear whether you are using permutations or combinations.
If you want to use permutations, $Pr =\dfrac{4!\cdot4}{7\cdot6\cdot5\cdot4}$
whereas if you want to use combinations, $Pr = \dfrac{4}{\binom7 4}$
Whenever probability is asked for, it is generally simpler to use combinations