Here's the problem:
Five couples sit randomly in a row during a movie. What is the probability that at least four couples sit next to each other.
My denominator is $10!$
The number of ways for all five couples to sit next to each other is $5!\cdot2^5$.
The number of ways for exactly four couples to sit next to each other is $(6!-5!\cdot2)\cdot2^4\cdot5$. The $6!-5!\cdot2$ is the number of ways for two specific people and four couples to arrange themselves minus the number of ways the two specific sit next to each other. Then I multiplied by $2^4$ to rearrange the people in the four couples. Then I multiplied by five to account for which of the five couples is split up.
Can someone check my logic? My class wants me to double-check with experts because I'm not sure about my answer.
Best Answer
Your solution is correct.
Here is another approach to the case in which exactly four couples sit next to each other.
We choose which of the five couples is separated. If we initially view each of the other four couples as a single object, they can be arranged in a row in $4!$ ways. This creates five spaces, three between successive couples and two at the ends of the row. $$\square c \square c \square c \square c \square$$ To separate the members of the fifth couple, we choose two of these five spaces for the couple, then arrange them within these spaces. The spaces can be selected in $\binom{5}{2}$ ways. The members of the couple can be arranged within these spaces in $2!$ ways. The members of the couples that sit together can each be arranged in $2!$ ways. Hence, the number of seating arrangements in which exactly four couples sit together is $$5 \cdot 4!\binom{5}{2}2!^5 = 5!\binom{5}{2}2!^5 = (6! - 5!2!)\cdot 2^4 \cdot 5$$ as you can verify.
Therefore, the desired probability is $$\frac{5!\binom{5}{2}2!^5 + 5! \cdot 2^5}{10!}$$